Force & Translational Dynamics.
Unit 1 described motion. This unit asks the question that changed physics forever: why do things move the way they do? The answer is three laws written by a 24-year-old in 1665. They still rule the macroscopic universe today.
Why things move: the single idea that built mechanical physics
Isaac Newton was 24 years old, home from Cambridge because the plague had closed the university, when he spent eighteen months in the countryside thinking harder than almost anyone ever has. By the end of it he had invented calculus, derived the laws of motion, explained planetary orbits, and understood that the same force pulling an apple to the ground was what held the moon in its orbit. He was, at that point, the world's greatest living scientist. He was also, at that point, younger than most grad students.
What he figured out — his three laws of motion — is the subject of this unit. Unit 1 gave us the language to describe motion (position, velocity, acceleration). Unit 2 gives us the mechanism that causes motion to change. Together, these two units are the foundation of what physicists call "classical mechanics," and they explain the behavior of every macroscopic object in the universe — from a baseball to a spacecraft to a collapsing star.
What is a force?
A force is an interaction between two objects that can change motion. Two words matter here: interaction and change. Interaction means a force is always exerted by something on something else — you can't have a force without two objects. A single object alone in empty space has no forces on it. Change means a force is not what keeps something moving; it's what makes something's velocity change. A hockey puck sliding on frictionless ice needs no force to keep going. It needs a force only if it's speeding up, slowing down, or changing direction.
Forces are vectors. They have magnitude (measured in newtons, N) and direction. When multiple forces act on an object, we add them as vectors to get the net force:
The common forces in AP Physics 1
You will see these six forces again and again. Memorize what causes each, what direction it points, and how to compute its magnitude.
| Force | Cause | Direction | Magnitude |
|---|---|---|---|
| Weight (gravity) | Earth pulling on mass | Straight down | \(W = mg\) |
| Normal force | Surface supporting an object | Perpendicular to surface | Whatever's needed |
| Tension | Rope/string pulling on object | Along rope, away from object | Whatever's needed |
| Friction (static) | Surface resisting motion | Opposing attempted motion | \(f_s \leq \mu_s N\) |
| Friction (kinetic) | Sliding contact | Opposing actual motion | \(f_k = \mu_k N\) |
| Spring force | Spring displaced from equilibrium | Toward equilibrium | \(F_s = kx\) |
Note the "whatever's needed" entries for normal and tension. These forces adjust their magnitudes to whatever the situation requires. A normal force grows as big as it needs to be to prevent an object from sinking through a floor. A rope's tension grows to whatever value keeps the rope from breaking or extending (up to its breaking point). You don't compute these forces directly — you solve Newton's second law and back them out.
Free-body diagrams: the universal language of dynamics
Every Unit 2 problem starts with a free-body diagram — a simplified drawing showing just one object and all the forces acting on it. Draw the object as a dot. From the dot, draw one arrow for every force, pointing in the direction of that force, with length roughly proportional to its magnitude. Label each arrow.
(1) Show only forces on the chosen object, never forces that object exerts on others. (2) Draw each force as a separate straight arrow from the dot. (3) Don't draw components — show only the actual forces. (4) If two forces point the same direction (e.g., two ropes pulling right), draw them side by side, not on top of each other.
Good FBDs make problems easy. Bad FBDs make them impossible. The most common error is forgetting that gravity always acts — even if the object is on a table, even if the object is in the air, even if the problem doesn't mention it. Earth pulls on every massive object, every time.
Newton's first law: equilibrium
If the net force on an object is zero, its velocity is constant. This includes the special case of being at rest (\(v = 0\)), but it also includes moving at constant velocity. The first law overturned 2000 years of Aristotelian physics, which held that moving objects naturally slow down. Aristotle had never seen a frictionless surface. On one, objects just keep going.
When an object is at rest or moving at constant velocity, we say it's in translational equilibrium. Equilibrium problems are among the most common in Unit 2: a book sitting on a table, a block hanging from two ropes, a crate on an incline that isn't sliding. For any of these: write down all forces, resolve into components, set each sum to zero, solve.
A box hanging from two ropes
A 10 kg box hangs in equilibrium from two ropes. Rope A makes a 30° angle with the ceiling; rope B makes a 60° angle. Find the tensions. Use \(g = 10\) m/s².
Set up
Three forces on the box: weight \(W = mg = 100\) N downward, tension \(T_A\) along rope A, tension \(T_B\) along rope B. Resolve tensions into horizontal and vertical components.
Apply \(\sum F_x = 0\) and \(\sum F_y = 0\)
\(\sum F_x: \; T_B\cos 60° - T_A\cos 30° = 0 \Rightarrow T_B(0.5) = T_A(0.866) \Rightarrow T_B = 1.732 T_A\)
\(\sum F_y: \; T_A\sin 30° + T_B\sin 60° - 100 = 0 \Rightarrow 0.5 T_A + 0.866 T_B = 100\)
Substitute: \(0.5 T_A + 0.866(1.732 T_A) = 100 \Rightarrow 0.5 T_A + 1.5 T_A = 100 \Rightarrow T_A = 50\) N.
Then \(T_B = 1.732(50) = 86.6\) N.
\(T_A = 50\) N; \(T_B = 86.6\) N.
Newton's second law: the workhorse
When the net force on an object is not zero, the object accelerates:
This is the single most important equation in all of physics. Two things to notice. First, it's a vector equation — it's really three equations in disguise (\(F_x = ma_x\), \(F_y = ma_y\), and so on). Second, acceleration points in the same direction as the net force, not the velocity. An object slowing down has acceleration pointing opposite its velocity, because the net force is backward.
(1) Draw a free-body diagram. (2) Choose coordinates — usually \(x\) along the direction of motion, \(y\) perpendicular. If on an incline, align one axis with the slope. (3) Write \(\sum F_x = ma_x\) and \(\sum F_y = ma_y\). (4) Substitute known forces. (5) Solve for unknowns. This recipe works for every Unit 2 problem without exception.
A block sliding down a ramp
A 5 kg block slides down a 30° incline with coefficient of kinetic friction 0.2. Find the block's acceleration. Use \(g = 10\) m/s².
Choose coordinates along and perpendicular to the incline
Let \(x\) point down the slope; \(y\) point perpendicular out of the slope.
Forces: weight \(mg\) (vertical) splits into \(mg\sin\theta\) down-slope and \(mg\cos\theta\) into-slope. Normal force \(N\) perpendicular-out. Kinetic friction \(f_k = \mu_k N\) up the slope (opposing motion).
Newton's second law
\(\sum F_y = 0: \; N - mg\cos\theta = 0 \Rightarrow N = mg\cos\theta = 50(0.866) = 43.3\) N.
\(\sum F_x = ma: \; mg\sin\theta - \mu_k N = ma\). Substitute: \(50(0.5) - 0.2(43.3) = 5a \Rightarrow 25 - 8.66 = 5a \Rightarrow a = 3.27\) m/s².
\(a \approx 3.3\) m/s² down the slope.
Newton's third law: every force has a partner
If object A exerts a force on object B, then object B exerts an equal and opposite force on object A. Always. No exceptions.
These are called Newton's third law pairs. They're always equal in magnitude, opposite in direction, and — this is the key — act on different objects. When Earth pulls down on you with force \(mg\), you pull up on Earth with force \(mg\). The Earth doesn't visibly accelerate because its mass is enormous, but the force is there. It's not zero.
Be careful not to confuse third-law pairs with balanced forces. When a book sits on a table, the normal force from the table and the weight of the book are both on the book — they balance, giving zero net force. But they're NOT third-law pairs. Third-law pairs act on different objects. The partner of the normal force (table pushing book up) is the force the book exerts on the table (book pushing table down). Third-law partners always sit on different free-body diagrams.
Gravitational force: weight from Earth
Near Earth's surface, every object experiences gravity pulling it toward Earth's center. We call this force the object's weight:
Weight depends on location (different \(g\) on different planets); mass doesn't. A 60 kg astronaut has mass 60 kg everywhere in the universe, but weighs 600 N on Earth, 100 N on the Moon, and 0 N in deep space.
The more general form of gravity — Newton's law of universal gravitation — is:
This gives the gravitational force between any two masses separated by distance \(r\). \(G = 6.67 \times 10^{-11}\) N·m²/kg² is the gravitational constant. Near Earth's surface, \(F_g = mg\) is just a special case with \(g = GM_E/R_E^2 \approx 9.8\) m/s². Universal gravitation comes up again in Unit 6 for orbits.
Friction: static vs kinetic
Two surfaces in contact resist sliding. When nothing is sliding, static friction adjusts itself to prevent motion, up to a maximum:
This is the most misunderstood formula in AP Physics 1. Static friction is whatever's needed to prevent motion, as long as it's less than \(\mu_s N\). If you push a heavy box and it doesn't move, static friction equals your push. If you push harder, static friction grows to match. Only when you exceed \(\mu_s N\) does the box start to slide.
Once sliding begins, friction becomes kinetic, with a fixed (and usually smaller) magnitude:
\(\mu_k\) is almost always less than \(\mu_s\) for the same surfaces. That's why it's harder to start a heavy box sliding than to keep it going once it's moving.
Springs: Hooke's law
A spring, when stretched or compressed by a distance \(x\) from its natural length, exerts a restoring force:
Here \(k\) is the spring constant (measured in N/m), and the minus sign indicates the force pulls the spring back toward equilibrium. The linear relationship between displacement and force is called Hooke's law — it's an approximation that's accurate for small displacements. Springs appear again in Unit 7 (oscillations).
Uniform circular motion: centripetal force
An object moving in a circle at constant speed is accelerating — its velocity direction is constantly changing, even though its speed isn't. The direction of this acceleration is always toward the center of the circle:
By Newton's second law, this requires a net inward force:
This is the centripetal force — but it's not a new kind of force. It's just a name for whatever force (or combination of forces) happens to be providing the net inward force required for circular motion. For a car turning, it's friction. For a planet orbiting, it's gravity. For a ball on a string swung in a circle, it's tension. The centripetal force is a requirement, not a cause — you solve circular motion problems by identifying which real forces are providing the required inward net force.
A car rounding a flat curve
A 1000 kg car rounds a flat curve of radius 50 m at 20 m/s. What is the minimum coefficient of static friction between tires and road that allows this turn without sliding?
Identify the required centripetal force
Required inward force: \(F_c = mv^2/r = (1000)(400)/50 = 8000\) N.
Source of inward force
On a flat road, only friction can point toward the center. So \(f_s \geq 8000\) N.
Apply max friction condition
Max static friction is \(\mu_s N = \mu_s mg = \mu_s (10{,}000)\). Require \(\mu_s(10{,}000) \geq 8000\), so \(\mu_s \geq 0.8\).
\(\mu_s \geq 0.8\). (This is actually higher than the friction of tires on wet roads — which is why cars slide on wet curves.)
Systems and center of mass
When multiple objects are connected (two blocks joined by a rope, a chain of masses on a track), you can treat them as a system and apply Newton's laws to the whole thing at once. The external forces on the system determine the acceleration of the system's center of mass:
Internal forces (like the tension between two connected blocks) cancel out in this equation and don't affect the center-of-mass motion. This is why you can ignore internal forces for the overall motion but need to consider them when solving for individual blocks.
Conceptual summary
Force is an interaction between objects, measured as a vector in newtons. Newton's three laws are the rules: objects with zero net force have constant velocity (1st), objects with nonzero net force accelerate at \(a = F_{\text{net}}/m\) (2nd), and every force has an equal-and-opposite partner on a different object (3rd). Common forces include gravity (\(mg\)), normal (whatever is needed, perpendicular to surfaces), tension (along ropes), friction (static up to \(\mu_s N\), kinetic equal to \(\mu_k N\)), and spring (\(-kx\)). Every problem starts with a free-body diagram. Uniform circular motion requires a net inward force of \(mv^2/r\), provided by whatever forces happen to be pointing inward. Choosing good coordinates (aligned with motion or along an incline) is the difference between hard algebra and easy algebra.
Every equation in Unit 2
★ Newton's second law
- \(\vec F_{\text{net}}\)
- vector sum of all forces on object (N)
- \(m\)
- mass (kg)
- \(\vec a\)
- acceleration vector (m/s²)
Use when object's velocity is changing. The workhorse of Unit 2.
★ Translational equilibrium
- Condition
- object at rest or constant velocity
Use when \(a = 0\). Often split into \(\sum F_x = 0\) and \(\sum F_y = 0\).
★ Weight
- \(g\)
- ≈ 9.8 m/s² near Earth's surface (use 10 for quick estimates)
Use when gravity is relevant. Always points toward Earth's center (down).
★ Universal gravitation
- \(G\)
- 6.67 × 10⁻¹¹ N·m²/kg²
- \(r\)
- center-to-center distance
Use when objects are far from Earth's surface, in orbit, or comparing planets.
★ Static friction
- \(\mu_s\)
- coefficient of static friction
- \(N\)
- normal force (not mg)
Use when object is not sliding. Take equality only at the point of slipping.
★ Kinetic friction
- \(\mu_k\)
- coefficient of kinetic friction
Use when object is sliding. Opposes direction of motion. Usually \(\mu_k < \mu_s\).
★ Spring force (Hooke's law)
- \(k\)
- spring constant (N/m)
- \(x\)
- displacement from equilibrium
Use when a spring is stretched or compressed. Minus = restoring.
★ Centripetal acceleration
- \(v\)
- speed around circle
- \(r\)
- radius of circle
Use when circular motion. Always points toward center.
★ Net force for circular motion
- Required net force
- toward center
Use when solving circular motion. Identify what real forces provide this net inward force.
Newton's third law
- Pairs
- act on different objects
Use when identifying force pairs or analyzing systems of interacting objects.
One page. Night before the test.
The 6 concepts that matter most
- Every problem starts with a free-body diagram. Forces on the object only, as arrows from a single dot, no components, labeled.
- Newton's second law is \(F_{\text{net}} = ma\) — a vector equation. Split into \(x\) and \(y\) components, solve each separately.
- Choose coordinates to match the motion. On an incline, tilt the axes so one points along the slope. This turns nasty trig into simple algebra.
- Static friction is "whatever's needed" up to the limit \(\mu_s N\). Only use \(f_s = \mu_s N\) at the point of slipping, not always.
- Centripetal force is a requirement, not a force. Identify which real forces (tension, gravity, friction, normal) happen to point inward — their sum must equal \(mv^2/r\).
- Third-law pairs act on different objects. When a book sits on a table, the normal force and the weight are both on the book — these balance but are NOT third-law pairs.
Key equations
- \(\vec F_{\text{net}} = m\vec a\)
- \(\sum \vec F = 0\) (equilibrium)
- \(W = mg\)
- \(f_s \leq \mu_s N\), \(f_k = \mu_k N\)
- \(F_s = -kx\) (spring)
- \(a_c = v^2/r\), \(F_{\text{net,in}} = mv^2/r\)
- \(F_g = Gm_1 m_2/r^2\)
6 traps students fall for
- Forgetting gravity — ALWAYS include \(mg\) on the FBD.
- Writing \(N = mg\) when object is on an incline (actually \(N = mg\cos\theta\)).
- Writing \(f_s = \mu_s N\) always (it's \(\leq\), not \(=\), unless sliding is imminent).
- Treating centripetal force as a separate force added to the FBD — it's not.
- Confusing third-law pairs with balanced forces on the same object.
- Forgetting that two connected objects must have the same magnitude of acceleration.
Problem-solving checklist
- Identify the object(s). Sometimes you need a separate FBD for each.
- Draw the FBD. Gravity first. Then any surface contact forces (normal, friction). Then any ropes (tension). Then applied forces.
- Choose coordinates. Align with motion or along an incline.
- Write Newton's second law in components. \(\sum F_x = ma_x\), \(\sum F_y = ma_y\).
- Solve the system. Substitute known values, eliminate unknowns, find what's asked.
- Sanity check. Units correct? Signs correct? Magnitude reasonable?
Ten problems building from basic to moderate
A 5 kg block rests on a horizontal surface. Find (a) the normal force, (b) the block's weight.
A 3 kg object experiences a net force of 12 N to the right. Find its acceleration.
A 10 kg box is pushed horizontally with 50 N of force across a floor with coefficient of kinetic friction 0.3. Find the box's acceleration.
A 2 kg block sits on a 30° incline with coefficient of static friction 0.4. Does the block slide? If not, what is the friction force? If yes, what is its acceleration down the slope (using \(\mu_k = 0.3\))?
Two blocks are connected by a light rope over a frictionless pulley. Block A (mass 3 kg) hangs vertically; block B (mass 2 kg) sits on a frictionless horizontal table. Find (a) the acceleration of the system, (b) the tension in the rope.
A spring has spring constant 200 N/m. A 4 kg mass is hung from it. How much does the spring stretch from its natural length?
A 1500 kg car drives around a flat, circular curve of radius 40 m at 15 m/s. What minimum coefficient of static friction between tires and road is needed to prevent the car from sliding?
A 0.5 kg ball is attached to a 1.5 m string and swung in a horizontal circle at 4 m/s. Find the tension in the string.
A 70 kg person stands in an elevator. Find the apparent weight (the normal force on the person's feet from the floor) when the elevator is (a) at rest, (b) accelerating upward at 2 m/s², (c) in free fall.
A 5 kg block is placed on a 20° incline and connected by a light rope over a frictionless pulley at the top to a 3 kg hanging block. The incline is frictionless. Find the system's acceleration and the direction of motion.
Ten problems in AP exam format
Part I — Multiple Choice
A book rests on a table. Which of the following pairs of forces forms a Newton's third-law action-reaction pair?
- The book's weight and the normal force from the table on the book.
- The force the book exerts on the table and the normal force the table exerts on the book.
- The book's weight and the force Earth exerts on the table.
- The normal force on the book and the book's acceleration.
A 5 kg block slides on a horizontal surface with coefficient of kinetic friction 0.2. A horizontal force of 30 N is applied. The acceleration of the block is closest to:
- 2 m/s²
- 4 m/s²
- 6 m/s²
- 10 m/s²
Two objects, one of mass \(m\) and one of mass \(2m\), are attached to the ends of a light string that passes over a massless frictionless pulley. If the system is released from rest, the acceleration of each object has magnitude:
- \(g/3\)
- \(g/2\)
- \(2g/3\)
- \(g\)
A ball on a string is swung in a vertical circle at constant speed. At which point in the circle is the tension greatest?
- At the top of the circle.
- At the bottom of the circle.
- At the sides of the circle (horizontal).
- The tension is the same at all points because the speed is constant.
A block of mass \(m\) sits on an incline at angle \(\theta\). The coefficient of static friction is \(\mu_s\). What is the maximum angle \(\theta_{\max}\) at which the block will remain stationary?
- \(\tan\theta_{\max} = \mu_s\)
- \(\sin\theta_{\max} = \mu_s\)
- \(\cos\theta_{\max} = \mu_s\)
- \(\theta_{\max} = \mu_s \cdot 90°\)
Part II — Short Free Response
(~7 min) A 2 kg block is pulled along a horizontal surface by a rope making a 30° angle above the horizontal. The rope tension is 20 N, and the coefficient of kinetic friction is 0.2.
- Draw a free-body diagram for the block.
- Derive a symbolic expression for the normal force in terms of \(m\), \(g\), \(T\), and \(\theta\).
- Calculate the block's acceleration.
- At what angle \(\theta\) would the normal force equal zero (i.e., the block is lifted off the surface)? Derive a symbolic expression.
(~7 min) A 1200 kg car drives over a hill with a circular profile of radius 30 m. At the top of the hill, the car is moving at speed \(v\).
- Draw a free-body diagram for the car at the top of the hill.
- Derive an expression for the normal force on the car at the top of the hill in terms of \(m\), \(g\), \(v\), and \(r\).
- At what speed would the car lose contact with the road at the top of the hill? Calculate the numerical value.
(~7 min) Two blocks, mass \(m_1 = 4\) kg and \(m_2 = 6\) kg, are pushed together across a horizontal frictionless surface by a force \(F = 20\) N applied to the left side of \(m_1\). They accelerate together.
- Find the acceleration of the two-block system.
- Find the contact force between the two blocks.
- If the force is instead applied to \(m_2\) (with \(m_1\) on the left), will the contact force be the same? Justify with a calculation.
Part III — Extended Free Response
(~18 min — experimental design) A student wants to determine the coefficient of static friction between a wooden block and a wooden board. The student has a protractor, the block, and a long wooden board. The student plans to tilt the board until the block just begins to slide, then measure the angle.
- Describe an experimental procedure to determine \(\mu_s\), including how to reduce experimental uncertainty.
- Derive an expression for \(\mu_s\) in terms of the critical angle \(\theta_c\) at which sliding begins.
- The student measures \(\theta_c = 25°\). Calculate \(\mu_s\).
- The student now replaces the wooden block with a different block of double the mass, on the same board. Does the critical angle change? Justify your answer using your expression from part (b).
- If the student instead measures the angle at which the block slides with constant velocity, how does that relate to \(\mu_k\)? Derive the expression.
(~18 min — modified Atwood machine) A block of mass \(m_1 = 3\) kg sits on a horizontal table with coefficient of kinetic friction \(\mu_k = 0.25\). It is connected by a light rope over a frictionless, massless pulley at the edge of the table to a hanging block of mass \(m_2 = 5\) kg.
- Draw separate free-body diagrams for both blocks.
- Derive a symbolic expression for the system's acceleration in terms of \(m_1, m_2, \mu_k,\) and \(g\).
- Calculate the numerical acceleration.
- Calculate the tension in the rope.
- If the coefficient of kinetic friction were doubled, would the system still accelerate? Justify.
Five problems at olympiad level
Block on block with friction. A 2 kg block sits on top of a 5 kg block on a frictionless horizontal surface. Coefficient of static friction between the blocks is 0.4. A horizontal force \(F\) is applied to the bottom block. Find the maximum \(F\) for which the two blocks move together without the top one sliding.
Banked curve. A car rounds a banked turn of radius \(r\) and bank angle \(\theta\). (a) Find the speed at which no friction is needed to prevent sliding. (b) Show that if the car travels faster than this, friction must point down the slope of the bank. Derive the maximum speed in terms of \(r\), \(\theta\), \(g\), and \(\mu_s\).
Conical pendulum. A mass \(m\) hangs from a string of length \(L\) and swings in a horizontal circle, with the string making a constant angle \(\theta\) with the vertical. (a) Derive the period of this motion in terms of \(L\), \(\theta\), and \(g\). (b) What happens to the period as \(\theta \to 90°\)? Does this make physical sense?
Three-block train. Three blocks of masses \(m\), \(2m\), and \(3m\) are connected in a row by two light ropes on a frictionless horizontal surface. A force \(F\) is applied to the front block (mass \(m\)). (a) Find the acceleration of the train. (b) Find the tension in each rope. (c) If instead the force is applied to the rear block (mass \(3m\)), do the tensions change? Derive expressions and compare.
Vertical loop minimum speed. A small object slides without friction along a track shaped like a vertical circular loop of radius \(R\). It enters the loop at the bottom with speed \(v_0\). (a) Find the minimum \(v_0\) required for the object to maintain contact at the top of the loop. (b) At this minimum speed, find the normal force at the bottom of the loop. Express in terms of \(m\), \(g\), and \(R\).
Every problem, every step
Worksheet A — Foundation
(a) Block in equilibrium. \(\sum F_y = 0: N - mg = 0 \Rightarrow N = mg = 50\) N. (b) \(W = mg = 50\) N.
\(a = F_{\text{net}}/m = 12/3 = 4\) m/s² to the right.
Normal: \(N = mg = 100\) N. Kinetic friction: \(f_k = \mu_k N = 0.3(100) = 30\) N, opposing motion.
Net force: \(F_{\text{net}} = 50 - 30 = 20\) N. Acceleration: \(a = 20/10 = 2\) m/s².
Force pulling down the slope: \(mg\sin\theta = 20(0.5) = 10\) N. Max static friction: \(\mu_s N = \mu_s mg\cos\theta = 0.4(20)(0.866) = 6.93\) N. Since \(10 > 6.93\), block slides.
Once sliding: kinetic friction \(f_k = \mu_k mg\cos\theta = 0.3(20)(0.866) = 5.20\) N. Net down-slope: \(10 - 5.20 = 4.80\) N. Acceleration: \(a = 4.80/2 = 2.4\) m/s² down.
Let \(a\) be the magnitude of acceleration, \(T\) the tension. For A (hanging, accelerating down): \(m_A g - T = m_A a\). For B (on frictionless table, accelerating horizontally): \(T = m_B a\).
Adding: \(m_A g = (m_A + m_B)a \Rightarrow a = 3(10)/5 = 6\) m/s². Tension: \(T = 2(6) = 12\) N.
Equilibrium: \(kx = mg \Rightarrow x = mg/k = 40/200 = 0.2\) m = 20 cm.
Centripetal force needed: \(F_c = mv^2/r = (1500)(225)/40 = 8438\) N. This must come from friction: \(\mu_s mg = \mu_s(1500)(10) = 15000\mu_s\). Require \(15000\mu_s \geq 8438 \Rightarrow \mu_s \geq 0.56\).
Tension provides the centripetal force (assume string is horizontal — idealization). \(T = mv^2/r = (0.5)(16)/1.5 = 5.33\) N.
Apparent weight = normal force \(N\). Newton's second law: \(N - mg = ma \Rightarrow N = m(g + a)\).
(a) At rest (\(a = 0\)): \(N = mg = 700\) N. (b) Upward \(a = 2\): \(N = 70(12) = 840\) N — person feels heavier. (c) Free fall (\(a = -g\)): \(N = 70(0) = 0\) N — weightless.
Let \(a\) be the acceleration of the system, with positive = block sliding up the incline (hanging block falling). Incline block: \(T - m_1 g\sin\theta = m_1 a\). Hanging block: \(m_2 g - T = m_2 a\).
Adding: \(m_2 g - m_1 g\sin\theta = (m_1 + m_2)a\). Check sign: \(m_2 g = 30\) N; \(m_1 g\sin 20° = 50(0.342) = 17.1\) N. \(m_2 g > m_1 g\sin\theta\), so hanging block falls.
\(a = (30 - 17.1)/8 = 1.61\) m/s². The incline block moves up-slope; the hanging block falls.
Worksheet B — AP Exam Style
Third-law pairs involve two objects. The book pushes down on the table; the table pushes up on the book — that's the pair. The book's weight and the normal force are both on the book (balanced forces, not a third-law pair).
\(f_k = \mu_k mg = 0.2(50) = 10\) N. Net force: \(30 - 10 = 20\) N. \(a = 20/5 = 4\) m/s².
\(a = (2m - m)g/(m + 2m) = g/3\).
At bottom: tension minus gravity provides centripetal. \(T - mg = mv^2/r \Rightarrow T = mg + mv^2/r\). At top: tension plus gravity provides centripetal. \(T + mg = mv^2/r \Rightarrow T = mv^2/r - mg\). Bottom tension is larger by \(2mg\).
On an incline at angle \(\theta\), block barely stationary: \(mg\sin\theta = \mu_s mg\cos\theta \Rightarrow \tan\theta = \mu_s\).
(a) FBD: weight \(mg\) down, normal \(N\) up, tension \(T\) at 30° above horizontal, friction \(f_k\) backward.
(b) Vertical equilibrium: \(N + T\sin\theta - mg = 0 \Rightarrow N = mg - T\sin\theta\).
(c) \(N = 20 - 20(0.5) = 10\) N. \(f_k = \mu_k N = 0.2(10) = 2\) N. Horizontal: \(T\cos\theta - f_k = ma \Rightarrow 20(0.866) - 2 = 2a \Rightarrow a = 7.66\) m/s².
(d) \(N = 0\) when \(T\sin\theta = mg \Rightarrow \sin\theta = mg/T \Rightarrow \theta = \arcsin(mg/T)\).
(a) FBD: weight \(mg\) down, normal \(N\) up. (b) Net downward force = centripetal: \(mg - N = mv^2/r \Rightarrow N = m(g - v^2/r)\). (c) \(N = 0\) when \(v^2 = gr = 300 \Rightarrow v = \sqrt{300} \approx 17.3\) m/s.
(a) Treat as one 10 kg object: \(a = 20/10 = 2\) m/s². (b) For block \(m_2\) alone: contact force \(F_c = m_2 a = 6(2) = 12\) N. (c) If push is on \(m_2\) instead: system still \(a = 2\) m/s². Now for block \(m_1\) alone: \(F_c = m_1 a = 4(2) = 8\) N. Different contact force — the "pushing" block bears more of the total force.
(a) Procedure: place block on board, slowly tilt, measure angle at the instant block starts to slide. Repeat several times, average. Use smallest possible contact area to reduce edge effects; ensure surfaces are clean.
(b) At critical angle: \(mg\sin\theta_c = \mu_s mg\cos\theta_c \Rightarrow \mu_s = \tan\theta_c\).
(c) \(\mu_s = \tan 25° = 0.466\).
(d) Mass cancels in the expression \(\mu_s = \tan\theta_c\) — so the angle doesn't change. Double the mass, same critical angle. Counterintuitive but correct.
(e) If block slides at constant velocity: \(\sum F = 0\), so \(mg\sin\theta = \mu_k mg\cos\theta \Rightarrow \mu_k = \tan\theta\).
(a) Block 1 (on table): weight, normal, tension forward, friction back. Block 2 (hanging): weight down, tension up.
(b) For block 1: \(T - \mu_k m_1 g = m_1 a\). For block 2: \(m_2 g - T = m_2 a\). Adding: \(m_2 g - \mu_k m_1 g = (m_1 + m_2)a\).
$$a = \dfrac{(m_2 - \mu_k m_1)g}{m_1 + m_2}$$
(c) \(a = (5 - 0.25(3))(10)/(8) = (5 - 0.75)(10)/8 = 42.5/8 = 5.31\) m/s².
(d) From block 2: \(T = m_2(g - a) = 5(10 - 5.31) = 23.4\) N.
(e) If \(\mu_k = 0.5\): \(a = (5 - 0.5(3))(10)/8 = 3.5(10)/8 = 4.38\) m/s². Still positive — still accelerates. The condition for no motion is \(m_2 < \mu_k m_1\), i.e., \(\mu_k > m_2/m_1 = 5/3 = 1.67\), which would be an unrealistically high friction coefficient.
Worksheet C — Challenge
Top block's only horizontal force is friction from the bottom block. For the top to accelerate at \(a\), \(f_s = m_{\text{top}}a\). Max \(a\) before top slides: \(\mu_s m_{\text{top}}g = m_{\text{top}}a \Rightarrow a_{\max} = \mu_s g = 4\) m/s². Then max \(F\) applied to bottom block (total mass 7 kg): \(F_{\max} = (m_1 + m_2)a_{\max} = 7(4) = 28\) N.
(a) No friction: horizontal component of normal gives centripetal. \(N\sin\theta = mv^2/r\) and \(N\cos\theta = mg\). Divide: \(\tan\theta = v^2/(rg) \Rightarrow v = \sqrt{rg\tan\theta}\).
(b) If \(v > v_{\text{design}}\), car tends to slide outward; friction must point inward (down the slope). Components along the center-directed axis: \((N\sin\theta + f\cos\theta) = mv^2/r\); perpendicular: \((N\cos\theta - f\sin\theta) = mg\). With \(f = \mu_s N\) at max speed: algebra gives \(v_{\max} = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}\).
(a) Vertical: \(T\cos\theta = mg\). Horizontal (centripetal): \(T\sin\theta = mv^2/r\), where \(r = L\sin\theta\). Divide: \(\tan\theta = v^2/(rg) \Rightarrow v^2 = rg\tan\theta = gL\sin\theta\tan\theta\). Period: \(T_{\text{period}} = 2\pi r/v = 2\pi L\sin\theta/\sqrt{gL\sin\theta\tan\theta} = 2\pi\sqrt{L\cos\theta/g}\).
(b) As \(\theta \to 90°\), \(\cos\theta \to 0\), so period \(\to 0\). Physical interpretation: to sustain a nearly horizontal circle, the ball must orbit infinitely fast. Makes sense — gravity still pulls it down, and a nearly-horizontal string gives almost no vertical support.
(a) Total mass \(6m\); \(a = F/(6m)\).
(b) Push on front: rope between \(m\) and \(2m\) pulls \(2m + 3m = 5m\) behind it. \(T_1 = 5m \cdot a = 5F/6\). Rope between \(2m\) and \(3m\) pulls \(3m\) behind it. \(T_2 = 3m \cdot a = F/2\).
(c) Push on rear \(3m\): rope between \(3m\) and \(2m\) pulls \(m + 2m = 3m\) in front. \(T_2 = 3m \cdot a = F/2\). Rope between \(2m\) and \(m\) pulls \(m\) in front. \(T_1 = m \cdot a = F/6\).
Compare: \(T_1\) dramatically different (\(5F/6\) vs \(F/6\)); \(T_2\) the same. Pushing from behind puts less strain on the front rope.
(a) At top of loop, minimum: gravity alone provides centripetal. \(mg = mv_{\text{top}}^2/R \Rightarrow v_{\text{top}}^2 = gR\). By energy conservation from bottom to top: \(\tfrac{1}{2}mv_0^2 = \tfrac{1}{2}mv_{\text{top}}^2 + mg(2R)\). So \(v_0^2 = v_{\text{top}}^2 + 4gR = gR + 4gR = 5gR \Rightarrow v_0 = \sqrt{5gR}\).
(b) At the bottom with \(v = v_0\): \(N - mg = mv_0^2/R = 5mg\), so \(N = 6mg\).
Seven errors that cost students points on Unit 2
Writing \(N = mg\) on an incline
A 5 kg block sits on a 30° ramp. Find the normal force.
Normal force is always perpendicular to the surface. On an incline, only the cosine component of gravity pushes into the surface; the sine component pulls the block down the slope. \(N = mg\) only on horizontal surfaces.
A 10 kg block sits on a 45° incline. Find the normal force and the component of gravity along the incline.
Treating \(f_s = \mu_s N\) as always true
A 5 kg block sits on a horizontal surface (\(\mu_s = 0.4\)) with 10 N applied horizontally. Find friction.
Static friction is \(\leq \mu_s N\), not \(= \mu_s N\). The equality holds only at the instant of slipping. Before that, static friction is whatever's needed to keep the object stationary.
A 20 kg crate sits on a floor with \(\mu_s = 0.3\). You push horizontally with 40 N. Does it move? If not, what is the friction force?
Forgetting to include gravity on the FBD
A ball is thrown horizontally off a cliff. Draw the FBD at the instant of release.
Students conflate "the force that set the ball moving" with "the forces currently on the ball." Once an object leaves your hand, you're no longer exerting a force on it. The ball's motion continues because of its velocity (Newton's first law), not because of a residual "force" from the throw.
A hockey puck slides on frictionless ice at constant velocity. Draw its FBD. (Hint: only two forces.)
Confusing third-law pairs with balanced forces
A book rests on a table. The weight of the book and the normal force from the table are a Newton's third law pair. True or false?
"Equal and opposite" describes both Newton's 3rd-law pairs AND balanced forces in equilibrium. The distinction is which objects they act on. If forces act on the same object and cancel, they're balanced. If they act on different objects, they're 3rd-law pairs.
For each force, identify its third-law partner: (a) you standing on the ground — normal force pushing you up; (b) a horse pulling a cart forward via a rope.
Adding "centripetal force" to the FBD
A ball on a string swings in a horizontal circle. Draw the FBD.
"Centripetal force" is a requirement, not a cause. You add up the real forces (gravity, tension, friction, normal, etc.), and the sum must equal \(mv^2/r\) toward the center. Don't list "centripetal force" separately — it's already accounted for by the real forces.
A car rounds a banked frictionless curve. Draw the FBD. Identify which force (or component of a force) provides the centripetal force.
Using weight (mg) instead of normal force for friction
A 4 kg block is pushed down an incline at 20° with a force of 30 N along the slope. Find the kinetic friction force. (\(\mu_k = 0.3\))
Friction depends on the normal force, not the weight. On flat ground they happen to be equal, but on an incline or when additional forces press on (or lift) the object, they're different.
A 6 kg block sits on a horizontal surface with \(\mu_k = 0.2\). A rope pulls upward on the block at 30° with 40 N. Find the friction force while the block slides horizontally.
Missing that connected objects share the same acceleration magnitude
Two blocks connected by a rope: one hangs, one is on a table. Set up equations.
A light, inextensible rope guarantees equal-magnitude acceleration of its two ends. This is the silent constraint that makes pulley problems solvable. Without it, you'd have too many unknowns.
Three blocks are connected by ropes in a vertical chain: 2 kg, 3 kg, and 5 kg from top to bottom, with the top rope supporting the whole system. All hang at rest. Find the tension in each rope.
What I learned going through this unit — and from helping others through it
What every student struggles with (in order)
The #1 sticking point in Unit 2 is that students skip the free-body diagram. They try to "see" the solution directly, write down some equation, and end up with the wrong answer. Every time. The FBD takes 30 seconds and eliminates 80% of errors. Do it even when the problem seems trivial. It's not optional — it's the single most important habit in mechanics.
The #2 struggle is coordinates on an incline. Students leave the \(x\) axis horizontal and the \(y\) axis vertical, then try to deal with components of the normal force. Always tilt the coordinates so one axis is along the slope — the normal force becomes pure \(y\), friction becomes pure \(x\), and only gravity has components (and those are easy: \(mg\sin\theta\) down-slope, \(mg\cos\theta\) into-slope).
The #3 struggle is the three "what's special about circular motion?" misconceptions: (1) thinking there's a "centrifugal force" pushing outward — there isn't, (2) treating centripetal force as a new force to add to the FBD — it's not, (3) forgetting that circular motion at constant speed still has acceleration — it does, directed toward the center.
What separates a 4 from a 5
Instinct for which coordinates to use. Students who score 5 immediately tilt their axes on inclines, decompose curved motion into radial and tangential directions, and choose whatever coordinates minimize the algebra. Students who score 4 stick with \(x\)-horizontal \(y\)-vertical and bury themselves in trig. The coordinate choice is often the hardest decision in a problem — get it right and the algebra is trivial.
The question that reveals true understanding
"A book sits motionless on a table. Name every force acting on the book and every Newton's third-law partner for each." A student who gets this cleanly has the conceptual core of Unit 2. The forces on the book: gravity (Earth pulling down) and normal force (table pushing up). These balance. The 3rd-law partners: gravity's partner is the book pulling Earth up (acts on Earth, not the book); normal force's partner is the book pushing the table down (acts on the table, not the book). Notice: no 3rd-law partner acts on the book itself. That's the whole point.
Exam-day strategy
For every problem: (1) Identify the object. (2) Draw the FBD. (3) Choose coordinates. (4) Write components of Newton's second law. (5) Solve.
For inclines: Tilt your coordinates. Weight components are \(mg\sin\theta\) along slope, \(mg\cos\theta\) into slope. Normal force balances only the perpendicular component, so \(N = mg\cos\theta\) (minus any other perpendicular forces).
For connected objects (pulleys, trains): Use the system approach to find acceleration (treat everything as one mass), then go back to individual objects to find internal forces.
For circular motion: Draw the FBD first. Then identify which forces (or components) point toward the center. Sum them and set equal to \(mv^2/r\).
For friction: First decide if the object is sliding. If not, \(f_s\) adjusts to whatever prevents motion (up to \(\mu_s N\)). If yes, \(f_k = \mu_k N\).
A surprising real-world connection
When you're in a car that turns a corner, you feel a push outward — the "centrifugal force." This force doesn't actually exist in an inertial reference frame. What's happening: your body's inertia wants to keep going in a straight line, but the car is turning. The car's door pushes you inward (normal force). You feel that inward push as an outward pull by comparison with the car's frame. This is why physics defines forces only in inertial (non-accelerating) reference frames — the "centrifugal force" is a fictitious artifact of analyzing motion from within an accelerating reference frame. It feels real because your body is being accelerated inward, but the only real force is the inward one.
One piece of advice
If you can't solve a Unit 2 problem, it's almost always because you drew an incomplete free-body diagram. Go back. Find the force you missed. It's usually gravity, tension, or the component of something that isn't perpendicular to anything. Unit 2 becomes easy when FBDs become automatic. Drill 10 FBDs of varying scenarios, and suddenly the rest of the unit clicks into place.