Unit 2: Force and Translational Dynamics

Useful variables

• $\vec{F}$ = force (Units: newtons, $\text{N} = \text{kg}\cdot\text{m}/\text{s}^2$)
• $m$ = mass (Units: kilograms, $\text{kg}$)
• $\vec{a}$ = acceleration (Units: $\text{m}/\text{s}^2$)
• $g$ = gravitational field strength near Earth = $9.8\ \text{m}/\text{s}^2$
• $\vec{W}$ or $\vec{F}_g$ = weight / gravitational force
• $F_N$ or $N$ = normal force
• $T$ = tension
• $f_s$ = static friction
• $f_k$ = kinetic friction
• $\mu_s$ = coefficient of static friction
• $\mu_k$ = coefficient of kinetic friction
• $k$ = spring constant (Units: $\text{N/m}$)
• $r$ = radius of circular path
• $v$ = speed
• $\omega$ = angular speed (Units: $\text{rad/s}$)

---

The Basics of Forces

What is a force? A force is a push or pull done by or on an object. Forces act like vectors. When a force is applied through contact, it is known as a contact force (e.g. friction, normal force) and when a force is applied with no contact, it is called a non-contact force (e.g. gravity, E&M force). Forces are commonly denoted with $\vec F$ and have units of Newtons ($N$)

Newton’s Three Laws of Motion

In 1687, Newton formulated the three laws of motion. They work extremely well for ordinary macroscopic objects moving much slower than the speed of light. At quantum, relativistic, or very strong-gravity scales, Newtonian mechanics must be replaced or extended, but for AP purposes, Newtonian mechanics suffices.

Theorem (Newton’s Three Laws).

1. First Law: An object has constant velocity unless acted on by a nonzero net external force. If $\sum \vec{F} = 0$, then $\vec{a} = 0$ (the converse is true as well).
2. Second Law: The net external force equals mass times acceleration, $\sum \vec{F} = m\vec{a}$.
3. Third Law: If object $A$ exerts a force on object $B$, then object $B$ exerts an equal-magnitude, opposite-direction force on object $A$, $\vec{F}_{A\text{ on }B} = -\vec{F}_{B\text{ on }A}$.

Third-law forces act on different objects, so they never cancel for one object. They can cancel only when you treat both interacting objects as one system and the force pair becomes internal.

Free-body diagrams

A free-body diagram is a force diagram for one object or one chosen system. It is the MOST important thing to do when solving force problems. It should show only external forces acting on that object/system, not forces the object applies to something else. Forces are treated as vectors and can be composed accordingly. An example of a free-body diagram is shown below:

Problem-solving strategy

1. Choose the object or system.
2. Draw a dot or simple sketch.
3. Add one arrow for each external force (make sure to draw where the force originates from).
4. Choose axes, usually one axis parallel to the expected acceleration or along a surface.
5. Resolve angled forces into components using vector decomposition.
6. Write Newton’s second law separately for each direction.

For a particle in two dimensions, you should split up force into components using vector decomposition like for velocity:

$$\sum F_x = ma_x, \qquad \sum F_y = ma_y.$$

If acceleration is zero in one direction, the net force in that direction is zero even if forces are present (basically they cancel out each other).

Inertial and non-inertial frames

Newton’s laws have their simplest form in an inertial frame, a frame that is not accelerating. However, in an accelerating frame (e.g. a moving train), you may introduce a pseudo-force so Newton’s second law appears to work inside that frame. Just like how you deal with relative velocity or acceleration, for a frame accelerating with $\vec{a}_{\text{frame}}$, the pseudo-force on a mass $m$ is

$$\vec{F}_{\text{pseudo}} = -m\vec{a}_{\text{frame}}.$$

Pseudo-forces are not interaction forces and do not have third-law partners (since they don’t actually exist!). Think of it like this: if a bus suddenly accelerates forward, you feel thrown backward, but no mysterious object pushed you backward. Your body was trying to keep its original velocity while the bus floor moved forward underneath you.

---

Common types of forces

In AP Physics C, most force problems are built from a small set of common forces.

Weight

Weight is the gravitational force on an object. Near Earth’s surface,

$$\vec{F}_g = m\vec{g},$$

so its magnitude is

$$F_g = mg.$$

Weight points downward, toward Earth’s center. Note that mass is not weight: mass is an object’s inertia, while weight is a force caused by gravity. Weight is always drawn from the center of the object, and sometimes drawing out the whole object (as opposed to just a dot) is very important! For example, when the force of gravity passes through the corner of the object, it will start to topple!

Normal force

The normal force is a contact force perpendicular to a surface and opposes gravity. It adjusts to prevent objects from passing through each other, but it is not automatically equal to $mg$. For example, on an incline or in an accelerating elevator, the normal force differs from the object’s weight. Normally, you need to solve out force equations to get the normal force. The normal force acts along the entire surface of contact but is usually drawn out in the center of the contact plane.

A concrete case where $F_N \ne mg$: suppose you push down on a box resting on the floor with an extra downward force $P$ at some angle, or simply press straight down. The vertical equation with no vertical acceleration is $F_N - mg - P = 0$, so $F_N = mg + P > mg$. If instead you pull up on the box with a force $P$ (not enough to lift it), then $F_N = mg - P < mg$. The normal force only equals $mg$ in the special case of a horizontal surface with no other vertical forces and no vertical acceleration.

Warning

Always get $F_N$ from the perpendicular equation, never by assumption, because friction limits like $\mu_s F_N$ depend on it.

Tension

Tension is a pulling force transmitted by a rope, string, or cable. In an ideal world, strings are massless and inextensible, and pulleys are massless and frictionless. Under those assumptions, the tension is the same throughout a continuous string. Tension will always point away from an object along the direction of the string. If a string or pulley has mass, or if the pulley has rotational inertia, tension may differ on different sides. Those cases usually belong more naturally with rotational dynamics.

To solve for tension, draw a separate free-body diagram for each attached object and write its force equation. Then connect the accelerations using the rope’s fixed length. Equal tension and equal acceleration are different assumptions: a movable pulley can have the same tension in its supporting segments while the rope end moves twice as far as the pulley.

For two masses hanging over a fixed pulley, take both vertical coordinates positive downward. The variable rope lengths satisfy

$$y_1+y_2=\text{constant} \quad\Longrightarrow\quad v_1+v_2=0 \quad\Longrightarrow\quad a_1+a_2=0.$$

For a movable pulley supported by two vertical segments, let $y_p$ be its downward position and $y_e$ the downward position of the free end beyond a fixed redirecting pulley. Then

$$2y_p+y_e=\text{constant} \quad\Longrightarrow\quad 2a_p+a_e=0.$$

The free end therefore accelerates with twice the magnitude and opposite sign. Use this relation with the force equations, counting both supporting tensions on the movable pulley. These constraints apply while the rope is taut; a rope cannot push, and a negative calculated tension means the assumed taut-rope motion is impossible.

Friction

Friction is a contact force that acts parallel to a surface that opposes relative motion or impending relative motion and will always point opposite to the direction of motion (e.g. when you are moving down a ramp friction points upwards). There are two types: static and kinetic friction.

Static friction is friction that prevents an object from moving, and adjusts up to a maximum value:

$$0 \le f_s \le f_{s,\text{max}} = \mu_s F_N.$$

Kinetic friction is friction an object experiences while moving and has approximately constant magnitude:

$$f_k = \mu_k F_N.$$

$\mu_s$ and $\mu_k$ are the coefficient of static friction and kinetic friction, respectively. Usually these values will be given, and determining them (without any other information) requires experimentation. An important note is that static friction is not always equal to $\mu_sF_N$; that expression gives the maximum possible static friction before slipping begins. Usually $\mu_s > \mu_k$.

Spring force

For an ideal spring, Hooke’s law gives the force required for displacement:

$$\vec{F}_s = -k\vec{x},$$

where $\vec{x}$ is displacement from equilibrium. The negative sign means the spring force points opposite the displacement. This force is more relevant in Unit 7: Oscillations. The value $k$ is called the spring constant.

Drag and resistive forces

Air resistance and fluid drag is friction experienced by an object going through a medium (usually liquid/air). They are often ignored in AP mechanics unless specified. When included, drag points opposite velocity (air resistance is drag in air). Two most common models are

$$\vec{F}_d = -b\vec{v}$$

for low-speed linear drag, and

$$\vec{F}_d = c\vec{v}^2$$

for high-speed quadratic drag, as seen in kinematics. The model that you should use for a problem will usually be stated.

Essentially, drag acts like a friction force for falling objects. At some point, the amount of drag pulling up will equal the force of gravity pulling down, allowing an object to go at a constant velocity (terminal velocity). Usually, drag force can never exceed gravity (that’s why skydivers don’t just go back up after reaching terminal velocity!).

---

Solving Newton’s second law problems

Problem-solving strategy

1. Choose the object or system, then draw its free-body diagram. For acceleration, start with the whole system when its objects share an acceleration. Internal contact forces will cacel.
2. If relevant, for tension or a contact force, isolate an object or subsystem on which that force is external and match the mass to the chosen system.
3. Pick axes that simplify the motion (e.g. $x$/$y$, radial/centripetal, etc.).
4. Write $\sum F = ma$ along each axis. Keep the algebra symbolic until the constraints and signs are settled, then check the units and whether the result agrees with the assumed direction of acceleration. Always include pseudo-forces if necessary.
5. Connect objects with constraints, such as equal tensions for an ideal string.
6. Solve algebraically before substituting numbers.
7. Check the sign and units of the result.

Common mistakes

• Do not assume $T=mg$ or $F_N=mg$. Other forces and acceleration can change either value.
• Do not count both a force and its components in the same sum. Components replace the original vector in that equation.
• Newton’s third-law partners act on different objects. They cancel in a combined system only when both objects are included.

If the question asks only for acceleration of a connected system, the system approach is often faster. If the question asks for tension or contact force, individual free-body diagrams are usually required.

Example. A block of mass $m_1 = 3.0\ \text{kg}$ sits on a frictionless horizontal table. A light inextensible string runs from the block, over a frictionless pulley at the edge of the table, to a hanging block of mass $m_2 = 2.0\ \text{kg}$. Find the acceleration of the system and the tension in the string.

Both blocks share the same acceleration magnitude $a$ because the string is inextensible: as $m_2$ falls, $m_1$ slides forward by the same amount.

First, we can treat the blocks as one system. The only external force along the direction of motion is the weight of the hanging mass, $m_2 g$, and the moving mass is $m_1 + m_2$:

$$m_2 g = (m_1 + m_2)a \quad\Rightarrow\quad a = \frac{m_2 g}{m_1 + m_2} = \frac{(2.0)(9.8)}{5.0} = 3.9\ \text{m/s}^2.$$

Now find the tension, which requires an individual free-body diagram. For the block on the table (horizontal direction, frictionless):

$$T = m_1 a = (3.0)(3.9) = 11.8\ \text{N}.$$

As a check, write Newton’s second law for the hanging mass with down positive:

$$m_2 g - T = m_2 a \quad\Rightarrow\quad T = m_2(g - a) = (2.0)(9.8 - 3.9) = 11.8\ \text{N}.$$

Both routes agree. Notice the system method gave $a$ instantly, but the tension only appeared once we cut the system into individual diagrams: tension is an internal force, invisible to the system equation.

---

Inclined planes

Tip

Tip

Always decompose forces along axes perpendicular and parallel to the surface of the incline. Check the gravity components against a horizontal surface: at $\theta=0$, the down-slope component $mg\sin\theta$ vanishes and the perpendicular component $mg\cos\theta$ becomes $mg$.

If a bob hangs perpendicular to the ceiling of a car translating along a straight incline, with its ceiling parallel to the incline, tension has no component along the slope. If the bob stays fixed relative to the car, its force equations give the car’s downhill acceleration $a=g\sin\theta$ and the tension $T=mg\cos\theta$.

For a block on an incline of angle $\theta$, it is usually best to choose axes parallel and perpendicular to the plane. The weight decomposes into

$$F_{g,\parallel} = mg\sin\theta,$$

down the incline, and

$$F_{g,\perp} = mg\cos\theta,$$

into/perpendicular to the incline.

If there is no acceleration perpendicular to the surface and no other force has a perpendicular component,

$$F_N = mg\cos\theta.$$

For a frictionless incline, the acceleration down the plane is

$$a = g\sin\theta.$$

With friction, decide whether the block is moving or about to move. If it is moving, use kinetic friction. If it is at rest, static friction takes whatever value is needed up to $\mu_sF_N$. It is also helpful to use geometry/similar triangles to determine certain angles for vector decompositions. ALWAYS remember your normal force!

Example. A block of mass $m = 4.0\ \text{kg}$ rests on an incline at $\theta = 30^\circ$, released from rest. The coefficients are $\mu_s = 0.50$ and $\mu_k = 0.40$. Determine whether the block slides, and if so, find its acceleration.

First decide between static and kinetic friction by comparing the driving force to the maximum static friction. The component of gravity down the plane is

$$F_{g,\parallel} = mg\sin\theta = (4.0)(9.8)\sin 30^\circ = 19.6\ \text{N}.$$

The normal force comes from the perpendicular equation (no acceleration perpendicular to the surface):

$$F_N = mg\cos\theta = (4.0)(9.8)\cos 30^\circ = 33.9\ \text{N}.$$

The maximum static friction is

$$f_{s,\text{max}} = \mu_s F_N = (0.50)(33.9) = 17.0\ \text{N}.$$

Since the driving force $19.6\ \text{N}$ exceeds $f_{s,\text{max}} = 17.0\ \text{N}$, static friction cannot hold the block, so it slides. Now use kinetic friction, which acts up the plane (opposing the downhill motion):

$$f_k = \mu_k F_N = (0.40)(33.9) = 13.6\ \text{N}.$$

Newton’s second law along the incline (down the plane positive):

$$mg\sin\theta - f_k = ma,$$ $$a = g\sin\theta - \mu_k g\cos\theta.$$

Substitute numbers:

$$a = 9.8(\sin 30^\circ - 0.40\cos 30^\circ) = 9.8(0.500 - 0.346) = 1.5\ \text{m/s}^2.$$

The block accelerates down the plane at about $1.5\ \text{m/s}^2$. Had $f_{s,\text{max}}$ exceeded $19.6\ \text{N}$, the block would have stayed put with static friction equal to exactly $19.6\ \text{N}$, not $\mu_s F_N$.

Example. A block of mass $m$ slides uphill on an incline of angle $\theta$ with kinetic friction coefficient $\mu_k$. A rope pulls at angle $\alpha$ above the slope. Derive its acceleration in terms of tension $T$, assuming it remains in contact, and specialize to a rope parallel to the slope.

A taut rope pulls along its own length. If it points up the incline, its tension acts entirely parallel to the surface. Taking up the slope as positive, a block sliding upward satisfies

$$T-mg\sin\theta-\mu_kF_N=ma, \qquad F_N=mg\cos\theta.$$

If the block slides downward, friction points up the slope instead, so its sign reverses. The acceleration sign still follows your chosen axis, not necessarily the direction of motion.

If the rope makes an angle $\alpha$ above the incline, resolve tension too. With no perpendicular acceleration,

$$F_N+T\sin\alpha-mg\cos\theta=0,$$

so $F_N=mg\cos\theta-T\sin\alpha$ while contact is maintained. For upward sliding, the parallel equation becomes

$$T\cos\alpha-mg\sin\theta-\mu_kF_N=ma.$$

The rope now both pulls the block uphill and reduces the normal force, which also reduces kinetic friction.

Substituting the normal force and dividing by mass gives

$$a=\frac{T(\cos\alpha+\mu_k\sin\alpha)}{m} -g(\sin\theta+\mu_k\cos\theta).$$

For a parallel rope, set $\alpha=0$. The contact assumption requires $mg\cos\theta-T\sin\alpha\ge0$.

Example. A rope parallel to a $30^\circ$ incline pulls a $4.0\ \text{kg}$ block uphill. The block is sliding upward and accelerating uphill at $1.2\ \text{m/s}^2$. If $\mu_k=0.20$, find the tension.

The rope has no perpendicular component, so

$$F_N=mg\cos30^\circ=33.9\ \text{N}.$$

Friction acts downhill because the block slides uphill. Its magnitude is $f_k=0.20(33.9)=6.79\ \text{N}$. Taking uphill as positive gives

$$T-mg\sin30^\circ-f_k=ma.$$

Therefore,

$$T=ma+mg\sin30^\circ+f_k =(4.0)(1.2)+19.6+6.79 =31.2\ \text{N}.$$

The tension must overcome both downhill forces and leave a net uphill force of $ma=4.8\ \text{N}$.

Friction and the slipping condition

Tip

Tip

For “will it slip?” questions, calculate the friction required to prevent relative sliding, then compare its magnitude with $\mu_sF_N$. If it exceeds that maximum, the assumed no-slip motion is impossible. Once the surfaces slide relative to each other, use $f_k=\mu_kF_N$.

Common mistakes

• Use $f_s=\mu_sF_N$ only at impending slip. Otherwise, static friction adjusts to the required value.
• Friction opposes relative sliding or the tendency to slide at the contact surface; it need not oppose the object’s ground-frame velocity.
• Recalculate $F_N$ when a force pulls away from or pushes into the surface. On a horizontal surface with zero vertical acceleration, an upward pull $T\sin\theta$ gives $F_N=mg-T\sin\theta$, provided contact is maintained.

The condition for impending slip (equivalent to when a block starts to move) is

$$f_s = \mu_s F_N,$$

equivalent to the maximum possible value for static friction. For a block resting on an incline, slipping begins when

$$mg\sin\theta = \mu_s mg\cos\theta.$$

Thus,

$$\tan\theta_{\max} = \mu_s.$$

This is the maximum angle before sliding for a simple block on a rough incline (rough meaning that there is friction). If external forces or other constraints are present, the slipping condition must be rederived from the free-body diagram.

The result $\tan\theta_{\max} = \mu_s$ is notable because the mass cancels: a heavy block and a light block of the same material begin to slide at the same angle. This is also a standard way to measure $\mu_s$ experimentally — slowly tilt a surface until the object just slips and record the angle.

Example. A coin placed on a flat book starts to slide when the book is tilted to $\theta_{\max} = 22^\circ$ from horizontal. Find $\mu_s$.

At impending slip the down-plane gravity component equals the maximum static friction:

$$mg\sin\theta_{\max} = \mu_s mg\cos\theta_{\max} \quad\Rightarrow\quad \mu_s = \tan\theta_{\max} = \tan 22^\circ \approx 0.40.$$

The mass and $g$ both cancel, which is why this simple tilt test works regardless of how heavy the coin is.

---

Connected objects and pulleys

Tip

Tip

For a massless rope passing over ideal massless, frictionless pulleys, tension is the same along each continuous rope. Different ropes can have different tensions. Count every segment pulling directly on your chosen system.

A movable pulley supported by two vertical segments feels an upward force $2T$. For a combined load and pulley of mass $M$, write $2T-Mg=Ma_y$, taking up as positive. Only when $a_y=0$ does this give $T=Mg/2$.

For a person pulling themselves up in a chair, count the rope’s pull on both the chair and the person’s hand. If each receives an upward tension $T$, the person-plus-chair system satisfies $2T-Mg=Ma_y$, where $M$ is their total mass.

For ideal ropes and pulleys, connected objects share related accelerations. A common Atwood machine is a very simple pulley that contains two hanging masses connected by a massless string over a frictionless pulley.

If $m_2 > m_1$, the acceleration magnitude is

$$a = \frac{(m_2 - m_1)g}{m_1 + m_2},$$

and the tension is

$$T = \frac{2m_1m_2}{m_1 + m_2}g.$$

Proof (Atwood tension and acceleration). Two masses $m_1$ and $m_2$ (with $m_2 > m_1$) hang from a massless, inextensible string over a massless, frictionless pulley. Find the acceleration $a$ and tension $T$.

Since the string is inextensible, whatever distance $m_2$ falls, $m_1$ rises by the same amount, so both masses have the same acceleration magnitude $a$. Take the direction of motion as positive for each mass: $m_2$ accelerates downward and $m_1$ accelerates upward, with the same $a$. The tension $T$ is the same on both sides because the string and pulley are ideal.

Free-body diagram for $m_1$ (taking up as positive):

$$T - m_1 g = m_1 a.$$

Free-body diagram for $m_2$ (taking down as positive):

$$m_2 g - T = m_2 a.$$

Add the two equations to eliminate $T$:

$$m_2 g - m_1 g = (m_1 + m_2)a,$$

so

$$a = \frac{(m_2 - m_1)g}{m_1 + m_2}.$$

To find $T$, substitute $a$ back into the first equation:

$$T = m_1(g + a) = m_1 g\left(1 + \frac{m_2 - m_1}{m_1 + m_2}\right) = m_1 g\cdot\frac{2m_2}{m_1 + m_2},$$

so

$$T = \frac{2m_1 m_2}{m_1 + m_2}g.$$

As a check, if $m_1 = m_2$ the acceleration is zero and $T = m_1 g$, as expected for balanced masses. If $m_2 \gg m_1$, then $a \to g$ (near free fall) and $T \to 2m_1 g$, never $m_2 g$.

For pulley systems with movable pulleys, the acceleration constraints may involve factors of 2. Write the string-length constraint (the length of the string is always constant) first, then differentiate with respect to time to relate velocities and accelerations.

---

Apparent weight and elevators

Common mistakes

• Apparent weightlessness means $F_N=0$, even while gravity acts.
• Use the direction of acceleration to decide whether the scale reading increases or decreases. An elevator traveling upward while slowing down accelerates downward.
• A negative calculated normal force means ordinary floor contact cannot persist: the floor cannot pull the person downward.

A scale does not directly measure your weight, it insteads measures the apparent weight, which is the normal force applied to the scale. In an elevator (or any condition with nonzero acceleration),

$$F_N - mg = ma$$

if upward is positive.

So,

$$F_N = m(g + a).$$

If the elevator accelerates upward, $F_N > mg$. If it accelerates downward, $F_N < mg$. In free fall, $a = -g$ and $F_N = 0$, so the object is weightless in the apparent-weight sense even though gravity still acts.

Example. A person of mass $m = 70\ \text{kg}$ stands on a bathroom scale in an elevator. Find the reading when (a) the elevator accelerates upward at $2.0\ \text{m/s}^2$, and (b) the elevator accelerates downward at $2.0\ \text{m/s}^2$.

Take up as positive. The scale reads $F_N$, where $F_N - mg = ma$, so $F_N = m(g + a)$.

(a) Upward acceleration, $a = +2.0\ \text{m/s}^2$:

$$F_N = 70(9.8 + 2.0) = 70(11.8) = 826\ \text{N}.$$

The person feels heavier than their true weight $mg = 686\ \text{N}$.

(b) Downward acceleration, $a = -2.0\ \text{m/s}^2$:

$$F_N = 70(9.8 - 2.0) = 70(7.8) = 546\ \text{N}.$$

The person feels lighter. Note the sign of $a$ is what matters, not the direction of motion: an elevator moving up but slowing down has $a < 0$ and gives the lighter reading.

---

Circular motion

Uniform circular motion

Kinematics tells us that an object moving in a circle needs an inward acceleration. Here, the new question is what real forces produce that acceleration. At each instant, choose a radial axis pointing toward the center and apply Newton’s second law along that axis:

$$\sum F_{\text{inward}}=ma_c=m\frac{v^2}{r}=m\omega^2r.$$

The expression $mv^2/r$ is not an additional force to draw on a free-body diagram. It is the required net inward force. Draw only forces that come from physical interactions, such as tension, gravity, friction, or a normal force, and then add their radial components.

Because the inward direction changes as the object moves, the signs of individual forces depend on the object’s location. A force pointing toward the center contributes positively to $\sum F_{\text{inward}}$; a force pointing away from the center contributes negatively; and a purely tangential force contributes zero to the radial equation.

Examples:

• A ball on a string: tension can provide the inward force.
• A car on a flat curve: static friction provides the inward force.
• A satellite in orbit: gravity provides the inward force.
• A roller coaster at the bottom of a loop: normal force and gravity combine to give the net inward force.

Problem-solving strategy

1. Draw a free-body diagram containing only real forces.
2. Mark the inward radial direction at the object’s current position.
3. Resolve each force into radial and tangential components.
4. Write $\sum F_{\text{inward}}=mv^2/r$. Use a separate tangential equation only if the speed is changing.

Example. A car takes a flat (unbanked) curve of radius $r = 50\ \text{m}$. The coefficient of static friction between the tires and road is $\mu_s = 0.60$. What is the maximum speed at which the car can round the curve without skidding?

On a flat road the only horizontal force available to turn the car is static friction, which must supply the centripetal force. At the maximum speed friction is at its limit:

$$\frac{mv_{\max}^2}{r} = \mu_s F_N = \mu_s mg.$$

The mass cancels, leaving

$$v_{\max} = \sqrt{\mu_s g r} = \sqrt{(0.60)(9.8)(50)} = \sqrt{294} \approx 17\ \text{m/s}.$$

That is about $62\ \text{km/h}$. Because $m$ cancels, a fully loaded truck and a light car can take the same curve at the same maximum speed (assuming equal $\mu_s$). Going faster than $v_{\max}$ means the required centripetal force exceeds what friction can supply, and the car slides outward.

Example. A ball of mass $m$ on a string of length $L$ swings in a horizontal circle, with the string tracing a cone at a constant angle $\theta$ from the vertical. Find the period $\mathcal{T}$ of the motion.

The ball moves in a horizontal circle of radius $r = L\sin\theta$, so its acceleration is purely horizontal and points toward the center. Two forces act: tension $T$ along the string and weight $mg$ down. There is no vertical acceleration, so the vertical components balance:

$$T\cos\theta = mg.$$

The horizontal component of tension provides the centripetal force:

$$T\sin\theta = \frac{mv^2}{r} = m\omega^2 r.$$

Divide the second equation by the first to eliminate $T$ and $m$:

$$\tan\theta = \frac{\omega^2 r}{g} = \frac{\omega^2 L\sin\theta}{g}.$$

Cancel $\sin\theta$ using $\tan\theta = \sin\theta/\cos\theta$:

$$\frac{1}{\cos\theta} = \frac{\omega^2 L}{g} \quad\Rightarrow\quad \omega = \sqrt{\frac{g}{L\cos\theta}}.$$

Since $\omega = 2\pi/\mathcal{T}$,

$$\mathcal{T} = 2\pi\sqrt{\frac{L\cos\theta}{g}}.$$

As $\theta \to 0$ the period approaches $2\pi\sqrt{L/g}$, the small-angle pendulum result. As $\theta \to 90^\circ$ the period goes to zero — you would need infinite tension to hold the string horizontal, which is why the string can never be perfectly horizontal.

Circular-motion tips and limiting conditions

Common mistakes

• Draw the inward direction before assigning signs. At the bottom of an inside loop, $F_N-mg=mv^2/r$; at its top, $F_N+mg=mv^2/r$. At the crest of a hill with the object on top of the track, $mg-F_N=mv^2/r$.
• The condition $F_N=0$ marks loss of ordinary contact, but the geometry decides whether it gives a maximum or minimum speed. At a hill’s crest it gives $v_{\max}=\sqrt{gr}$; at the top of an inside loop it gives $v_{\min}=\sqrt{gr}$. These assume the track can only push, with gravity and the normal force as the only radial forces.
• For changing speed, $v^2/r$ is only the radial acceleration. Also write $\sum F_t=ma_t$, and use $a=\sqrt{a_r^2+a_t^2}$ for the total magnitude.

Tip

Tip

For a pendulum with angle $\theta$ measured from the downward vertical, take inward as positive radially and increasing $\theta$ as positive tangentially. Then $T-mg\cos\theta=mv^2/L$ and $ma_t=-mg\sin\theta$. Tension has no tangential component. The minus sign means gravity accelerates the bob toward the bottom of the swing.

Horizontal circles and hanging objects

Tip

Tip

• Determine the actual path radius before substituting into $mv^2/r$. For a conical pendulum, $r=L\sin\theta$. For a swing attached at platform radius $R$ and tilted outward, $r=R+L\sin\theta$, where $\theta$ is measured from vertical.
• Divide the horizontal and vertical tension equations to eliminate tension. A bob fixed relative to a car with horizontal acceleration magnitude $a$ satisfies $T\sin\theta=ma$ and $T\cos\theta=mg$, so $\tan\theta=a/g$. On a level constant-speed turn, substitute $a=v^2/r$.
• In a rotating-cylinder ride, the wall’s normal force supplies the radial acceleration: $F_N=m\omega^2r$. Static friction supports the rider vertically, so $f_s=mg\le\mu_sF_N$. Thus $\omega_{\min}=\sqrt{g/(\mu_s r)}$. Convert angular speed to revolutions per second using $f=\omega/(2\pi)$.

Banked curves

Tip

Tip

Resolve forces horizontally toward the center and vertically. Even though the road is tilted, a car following a level circular path has no vertical acceleration. Determine whether it would slide up or down the bank without friction, then point static friction against that tendency.

Sometimes, circular motion is not confined on a flat surface. For a frictionless banked (raised) curve, the horizontal component of the normal force provides centripetal acceleration, while the vertical component balances weight:

$$F_N\cos\theta = mg,$$ $$F_N\sin\theta = \frac{mv^2}{r}.$$

Dividing gives

$$\tan\theta = \frac{v^2}{rg}.$$

So the design speed is

$$v = \sqrt{rg\tan\theta}.$$

With friction, static friction points whichever way prevents slipping: up the slope if the car would slide down, and down the slope if the car would slide up.

Proof (banked-curve design speed). A car rounds a curve of radius $r$ on a road banked at angle $\theta$, with no friction needed. Find the speed at which it can do so.

Only two forces act: the normal force $N$, perpendicular to the road surface, and the weight $mg$, straight down. Use horizontal and vertical axes (not axes along the incline), because the acceleration is horizontal — it points toward the center of the circle, which lies in the horizontal plane.

The normal force tilts inward by the bank angle $\theta$ from vertical. Resolve it: its vertical component is $N\cos\theta$ and its horizontal (inward) component is $N\sin\theta$.

Vertically, there is no acceleration, so the vertical forces balance:

$$N\cos\theta = mg.$$

Horizontally, the inward component of the normal force is the entire centripetal force:

$$N\sin\theta = \frac{mv^2}{r}.$$

Divide the horizontal equation by the vertical equation. Both $N$ and $m$ cancel:

$$\tan\theta = \frac{v^2}{rg}.$$

Solving for $v$ gives the design speed:

$$v = \sqrt{rg\tan\theta}.$$

At exactly this speed friction is not required at all. The mass cancels, so the design speed is the same for every vehicle.

Example. A highway curve of radius $r = 120\ \text{m}$ is to be banked so that a car traveling at $v = 25\ \text{m/s}$ (about $90\ \text{km/h}$) needs no friction. Find the required bank angle $\theta$.

From the design-speed relation,

$$\tan\theta = \frac{v^2}{rg} = \frac{(25)^2}{(120)(9.8)} = \frac{625}{1176} = 0.531,$$

so

$$\theta = \arctan(0.531) \approx 28^\circ.$$

A car going faster than $25\ \text{m/s}$ on this bank would tend to slide outward and up the slope, so static friction would point down the slope; a slower car would tend to slide inward and down, so friction would point up the slope. Friction therefore widens the safe range of speeds around the design speed.

Universal Gravitation

You may be familiar with the “force of gravity” on you, or in other words, your weight. However, $F=mg$ is a large simplification because gravity acts between two objects (by Newton’s third law), not just one, and we are assuming that the Earth experiences negligible force (and thus does not move) when compared with the weight you experience. In general, Newton’s law of universal gravitation gives the attractive force between two masses:

$$F_g=\frac{Gm_1m_2}{r^2}.$$

The force points along the line connecting the masses and both reactionary forces point towards each other. For a mass $m$ near a planet of mass $M$,

$$F_g=\frac{GMm}{r^2}=mg(r),$$

so the local gravitational field strength is

$$g(r)=\frac{GM}{r^2}.$$

Near Earth’s surface, $r\approx R_E$ changes very little over ordinary heights, so $g(r)$ is treated as constant and becomes the familiar $9.8\ \text{m/s}^2$. Far from the surface, however, you must use the inverse-square form.

Example. A satellite orbits Earth in a circular orbit of radius $r=7.0\times10^6\ \text{m}$ from Earth’s center. Using $GM_E=3.99\times10^{14}\ \text{m}^3/\text{s}^2$, find its orbital speed.

For a circular orbit, gravity supplies the centripetal force:

$$\frac{GM_Em}{r^2}=m\frac{v^2}{r}.$$

The satellite mass cancels, giving

$$v=\sqrt{\frac{GM_E}{r}} =\sqrt{\frac{3.99\times10^{14}}{7.0\times10^6}} =7.5\times10^3\ \text{m/s}.$$

The satellite is falling toward Earth, but its sideways speed is large enough that it keeps missing the surface.

Example. A planet of mass $M$ has a moon in a circular orbit of radius $r$ with measured orbital speed $v_{\text{orbit}}$. A second moon orbits the same planet at radius $4r$. Using only Newton’s law of universal gravitation and circular-motion force balance, find the second moon’s orbital speed in terms of $v_{\text{orbit}}$.

For the first moon, gravity supplies the centripetal force:

$$\frac{GMm}{r^2}=m\frac{v_{\text{orbit}}^2}{r}.$$

Cancel $m$ and solve for $GM$:

$$GM=v_{\text{orbit}}^2r.$$

For the second moon at radius $4r$,

$$\frac{GMm_2}{(4r)^2}=m_2\frac{v_2^2}{4r}.$$

Cancel $m_2$:

$$v_2^2=\frac{GM}{4r}.$$

Substitute $GM=v_{\text{orbit}}^2r$:

$$v_2^2=\frac{v_{\text{orbit}}^2r}{4r} =\frac{v_{\text{orbit}}^2}{4},$$

so

$$v_2=\frac{v_{\text{orbit}}}{2}.$$

The farther moon moves more slowly because the gravitational field is weaker and a larger orbit needs less centripetal acceleration for a given speed.

---

Variable forces and calculus form

For constant mass, Newton’s second law can be written as

$$\vec{F}_{\text{net}} = m\frac{d\vec{v}}{dt} = m\frac{d^2\vec{r}}{dt^2}.$$

For variable-mass systems, you will have to use methods that are learned later.

Example. An object of mass $m$ is released from rest and falls subject to gravity and linear drag $F_d = -bv$. Derive $v(t)$ and confirm the terminal velocity.

Take down as positive. Newton’s second law is

$$m\frac{dv}{dt} = mg - bv.$$

This is a separable first-order differential equation. Separate variables:

$$\frac{dv}{mg - bv} = \frac{dt}{m}.$$

Integrate the left side using $\int \frac{dv}{mg - bv} = -\frac{1}{b}\ln\lvert mg - bv \rvert$:

$$-\frac{1}{b}\ln(mg - bv) = \frac{t}{m} + C.$$

Apply the initial condition $v(0) = 0$ to find $C = -\frac{1}{b}\ln(mg)$. Substitute back and combine the logarithms:

$$-\frac{1}{b}\ln\!\left(\frac{mg - bv}{mg}\right) = \frac{t}{m}.$$

Multiply by $-b$ and exponentiate:

$$\frac{mg - bv}{mg} = e^{-bt/m}.$$

Solve for $v$:

$$v(t) = \frac{mg}{b}\left(1 - e^{-bt/m}\right).$$

Since the terminal velocity is $v_t = mg/b$, this is exactly

$$v(t) = v_t\left(1 - e^{-bt/m}\right).$$

As $t \to \infty$, the exponential vanishes and $v \to v_t$, as expected. The quantity $\tau = m/b$ is the time constant: after one time constant the speed reaches about $63\%$ of terminal velocity. At early times the exponential expands as $1 - bt/m + \dots$, giving $v \approx gt$, which matches a non-drag scenario.

---

Working checklist

Working checklist

1. Choose the system carefully.
2. Draw a clean free-body diagram.
3. Pick axes that match the geometry or acceleration.
4. Break forces into components.
5. Write $\sum F_x = ma_x$ and $\sum F_y = ma_y$.
6. Add constraints for ropes, pulleys, or circular motion.
7. Solve symbolically when possible.
8. Check limiting cases: frictionless, very large mass, zero angle, or zero acceleration.
9. For circular motion, point one axis along the radius (toward the center) and set the net radial force equal to $mv^2/r$.

---

Key equations

Idea	Equation
Newton’s second law	$\sum \vec{F} = m\vec{a}$
Components	$\sum F_x = ma_x,\quad \sum F_y = ma_y$
Weight near Earth	$F_g = mg$
Static friction	$0 \le f_s \le \mu_sF_N$
Kinetic friction	$f_k = \mu_kF_N$
Hooke’s law	$\vec{F}_s = -k\vec{x}$
Incline components	$mg\sin\theta,\quad mg\cos\theta$
Uniform circular acceleration	$a_c = v^2/r = \omega^2r$
Radial Newton’s second law	$\sum F_r = mv^2/r$
Frictionless banked curve	$v = \sqrt{rg\tan\theta}$
Flat-curve max speed	$v_{\max} = \sqrt{\mu_s g r}$
Atwood machine	$a = \dfrac{(m_2 - m_1)g}{m_1 + m_2},\quad T = \dfrac{2m_1 m_2}{m_1 + m_2}g$
Apparent weight (elevator)	$F_N = m(g + a)$
Top-of-loop minimum speed	$v_{\text{top}} = \sqrt{gr}$
Terminal velocity (linear drag)	$v_t = mg/b$
Center of mass dynamics	$\sum \vec{F}_{\text{ext}} = M\vec{a}_{\text{CM}}$
Fixed pulley, both coordinates downward	$y_1+y_2=\text{constant},\quad a_1+a_2=0$
Movable pulley, two supporting segments, downward coordinates	$2y_p+y_e=\text{constant},\quad 2a_p+a_e=0$
Two vertical supporting tensions (up positive)	$2T-Mg=Ma_y$
Incline, rope at angle $\alpha$ above slope, contact maintained	$F_N=mg\cos\theta-T\sin\alpha$
Sliding uphill, uphill positive	$T\cos\alpha-mg\sin\theta-\mu_kF_N=ma$
Impending slip, gravity-only incline	$\tan\theta_{\max}=\mu_s$
Tangential force equation	$\sum F_t=ma_t$
Conical pendulum ($\theta$ from vertical)	$T\cos\theta=mg,\quad T\sin\theta=mv^2/r,\quad r=L\sin\theta$
Hill crest, ordinary contact	$mg-F_N=mv^2/r,\quad v_{\max}=\sqrt{gr}$
Pendulum ($\theta$ from downward vertical)	$T-mg\cos\theta=mv^2/L,\quad ma_t=-mg\sin\theta$
Bob fixed in horizontally accelerating car	$\tan\theta=a/g$
Rotating cylinder, rider not slipping	$F_N=m\omega^2r,\quad mg\le\mu_sF_N,\quad \omega_{\min}=\sqrt{g/(\mu_s r)}$
Universal gravitation and field	$F_g=Gm_1m_2/r^2,\quad g(r)=GM/r^2$
Circular orbit about a dominant central mass	$v=\sqrt{GM/r}$
Constant-mass calculus form	$\vec F_{\mathrm{net}}=m\,d\vec v/dt=m\,d^2\vec r/dt^2$
Linear drag force, falling from rest (down positive)	$m\,dv/dt=mg-bv,\quad v(t)=\frac{mg}{b}(1-e^{-bt/m}),\quad \tau=m/b$

Practice

Multiple Choice

1. A block of mass $m$ rests on a small platform scale mounted on an incline of angle $\theta$. The wedge and scale are at rest, and static friction prevents slipping.

If the scale measures the normal force on the block, its reading is

(A) $mg\sin\theta$

(B) $mg\cos\theta$

(C) $mg\tan\theta$

(D) $mg$

2. A block is pressed against a vertical wall by a horizontal force $F$. The coefficient of static friction is $\mu_s$. The smallest $F$ that can keep the block from sliding is

(A) $mg$

(B) $\mu_s mg$

(C) $mg/\mu_s$

(D) $\mu_s/g$

3. Two blocks of masses $m$ and $2m$ are connected by a light string and pulled across a frictionless table by force $F$ applied to the $2m$ block. The tension in the string is

(A) $F/3$

(B) $F/2$

(C) $2F/3$

(D) $F$

4. A falling object experiences drag force $bv$ upward. Taking downward as positive, which differential equation describes the motion?

(A) $m\dfrac{dv}{dt}=mg+bv$

(B) $m\dfrac{dv}{dt}=mg-bv$

(C) $m\dfrac{dv}{dt}=bv-mg$

(D) $m\dfrac{dv}{dt}=-mg-bv$

5. A block of mass $m$ sits on a rough incline of angle $\theta$. A horizontal force $F$ pushes the block into the incline. Which change most directly increases the maximum possible static friction?

(A) Decreasing $F$

(B) Increasing $F$

(C) Decreasing $m$ while keeping $F$ fixed

(D) Making the incline frictionless

6. A pendulum bob hangs motionless relative to a train accelerating horizontally with magnitude $a$. If the string makes angle $\theta$ with the vertical and the tension is $T$, which pair of equations is consistent with the bob’s rest in the train’s frame?

(A) $T\sin\theta=ma$ and $T\cos\theta=mg$

(B) $T\cos\theta=ma$ and $T\sin\theta=mg$

(C) $T=mg$ and $\tan\theta=a/g$

(D) $T=ma$ and $\tan\theta=g/a$

7. A car travels over the top of a circular hill of radius $R$. At the top, the driver feels an apparent weight equal to one-third of their normal weight. The car’s speed is

(A) $\sqrt{gR/3}$

(B) $\sqrt{2gR/3}$

(C) $\sqrt{gR}$

(D) $\sqrt{4gR/3}$

8. A bead slides on a frictionless circular hoop in a vertical plane. At the side of the hoop, its speed is $v$. The normal force magnitude is

(A) $mg$

(B) $mv^2/R$

(C) $mg+mv^2/R$

(D) $\sqrt{(mg)^2+(mv^2/R)^2}$

9. An elevator accelerates upward with magnitude $a$. Inside it, a mass $m$ hangs from a spring scale while a horizontal force $F$ pulls the mass sideways so the supporting string makes angle $\phi$ with the vertical. The tension in the string is

(A) $m(g+a)$

(B) $\dfrac{m(g+a)}{\cos\phi}$

(C) $\dfrac{mg}{\cos\phi}$

(D) $m\sqrt{g^2+a^2}$

10. A block rests on a scale mounted on a wedge inclined at angle $\theta$. The wedge accelerates horizontally to the right with magnitude $a$, and the incline rises to the right. The block remains at rest relative to the scale. If the scale measures the normal force on the block, its reading is

(A) $m(g\cos\theta-a\sin\theta)$

(B) $m(g\cos\theta+a\sin\theta)$

(C) $m(g\sin\theta+a\cos\theta)$

(D) $m(g+a)\cos\theta$

11. A small mass moves in a vertical circle on a string of length $R$. Its speeds at the bottom and top are $v_b$ and $v_t$, and the corresponding string tensions are $T_b$ and $T_t$. Which relation follows from Newton’s second law in the radial direction?

(A) $T_b-T_t=\dfrac{m(v_b^2-v_t^2)}{R}+2mg$

(B) $T_b-T_t=\dfrac{m(v_b^2-v_t^2)}{R}$

(C) $T_b+T_t=\dfrac{m(v_b^2+v_t^2)}{R}$

(D) $T_b-T_t=2mg-\dfrac{m(v_b^2-v_t^2)}{R}$

12. A block of mass $m$ rests on the floor of an elevator that accelerates upward with magnitude $a_y$ while also accelerating horizontally with magnitude $a_x$. The block does not slip relative to the floor. The minimum coefficient of static friction required is

(A) $\dfrac{a_x}{g+a_y}$

(B) $\dfrac{a_x}{g}$

(C) $\dfrac{g+a_y}{a_x}$

(D) $\dfrac{\sqrt{a_x^2+(g+a_y)^2}}{g}$

FRQ

1. A block of mass $m$ is inside a box that accelerates horizontally with acceleration $a$. The block is pressed against the box’s vertical wall and does not slip. The coefficient of static friction between the block and wall is $\mu_s$.

   $(A)$ Draw a free-body diagram for the block in the ground frame.

   $(B)$ Derive the normal force exerted by the wall on the block.

   $(C)$ Determine the condition on $a$ for the block not to slide down.

   $(D)$ If the box also accelerates upward with acceleration $a_y$, derive the new no-slip condition.

2. A bead of mass $m$ slides without friction on a circular hoop of radius $R$ fixed in a vertical plane. At an angle $\theta$ measured from the bottom, the bead has speed $v$

   $(A)$ Draw a force diagram for the bead.

   $(B)$ Write Newton’s second law in the radial direction.

   $(C)$ Write Newton’s second law in the tangential direction.

   $(D)$ At angle $\theta$, determine the speed at which the bead would just lose contact with the hoop, if such a speed is possible.

3. A mass $m$ falls from rest through a fluid with drag force $F_d=bv$ upward. Take downward as positive.

   $(A)$ Write the differential equation for $v(t)$.

   $(B)$ Determine the terminal speed.

   $(C)$ Without solving fully for $v(t)$, determine whether the acceleration is increasing, decreasing, or constant as the object falls.

   $(D)$ Design a linear graph that could be used to determine $b$ from measurements of speed and acceleration.