Friction and Tension — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Static friction, kinetic friction, tension force, ideal ropes and pulleys, the friction coefficient formula, free-body diagram methods for tension-friction systems, and connected object analysis for AP Physics 1 dynamics problems.

You should already know: Newton's three laws of motion, how to draw and interpret free-body diagrams, vector resolution of force components.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Friction and Tension?

Friction is a contact force that opposes relative motion between two solid surfaces in contact, while tension is a pulling force transmitted through a flexible stretched medium (like a rope, string, or cable). Both are core subtopics in AP Physics 1 Unit 2: Dynamics, which accounts for 12–18% of the total AP exam score; friction and tension make up roughly a third of this unit weight, appearing regularly in both multiple-choice (MCQ) and free-response (FRQ) sections.

Standard notation: Friction is written as $f$, with subscripts $s$ for static friction and $k$ for kinetic friction; tension is almost always labeled $T$. Friction arises from microscopic irregularities of surfaces and intermolecular attractions between the two materials, while tension arises from electrostatic bonds between molecules in the stretched medium. For AP Physics 1, we almost always assume ideal ropes (massless, inextensible) and ideal pulleys (massless, frictionless) unless explicitly stated otherwise, which simplifies analysis because tension is uniform along an ideal rope.

2. Static and Kinetic Friction

Friction is split into two categories based on whether the surfaces are moving relative to each other. Static friction acts when there is no relative motion between the surfaces; it adjusts its magnitude to exactly oppose the parallel component of the applied force trying to move the object, up to a maximum threshold. The formula for maximum static friction is: $$f_{s,max}={\mu }_{s}N$$ where ${\mu }_s$ is the dimensionless coefficient of static friction (dependent on the two surface materials), and $N$ is the magnitude of the normal force perpendicular to the contact surface. Kinetic friction acts when the surfaces are sliding relative to each other; it has a constant magnitude given by: $$f_k={\mu }_{k}N$$ For any pair of surfaces, ${\mu }_k<{\mu }_s$, which means it takes more force to start moving an object than to keep it moving at constant speed. A common misconception is that normal force always equals an object’s weight; this is only true for horizontal surfaces with no additional vertical forces. $N$ must always be calculated from Newton’s second law in the direction perpendicular to the contact surface.

Worked Example

A 12 kg wooden crate rests on a horizontal concrete floor, with ${\mu }_s=0.6$ and ${\mu }_k=0.4$. What is the magnitude of friction when a horizontal 50 N force pushes on the stationary crate?

1. Calculate the normal force: No vertical acceleration, so $N=mg=12\times 9.8=117.6\text{ N}$.
2. Calculate maximum static friction: $f_{s,max}={\mu }_{s}N=0.6\times 117.6=70.56\text{ N}$.
3. Compare the applied force to the maximum threshold: $50\text{ N}<70.56\text{ N}$, so the crate remains stationary.
4. For stationary objects, static friction matches the applied parallel force: $f_s=50\text{ N}$.

Exam tip: Always compare the applied force to $f_{s,max}$ before assuming friction is kinetic. AP exam questions regularly trick students into automatically using $f_k$ when the object is not moving.

3. Tension in Ideal Ropes and Pulleys

Tension is a pulling force that acts along the length of a rope, pulling equally on both objects connected to the rope. For AP Physics 1, all ropes and pulleys are assumed to be ideal unless stated otherwise:

• An ideal rope is massless and inextensible: Inextensible means all objects connected by the rope have the same magnitude of acceleration, even if acceleration directions differ. Massless means the net force on the rope must be zero (from $F_{net}=ma$, $m=0$ so $F_{net}=0$), so tension is the same magnitude at both ends of the rope.
• An ideal fixed pulley is massless and frictionless: it only changes the direction of tension, not its magnitude, so tension remains equal on both sides of the pulley.

Worked Example

A 5 kg mass hangs vertically from an ideal rope that runs over a fixed ideal pulley, connected to an 8 kg block resting on a frictionless horizontal table. What is the magnitude of tension in the rope?

1. Assign acceleration: The hanging mass accelerates downward, the block accelerates to the right, with equal magnitude $a$.
2. Write Newton’s second law for each object: For the 8 kg block (horizontal direction): $\sum F=T=8a$. For the 5 kg hanging mass (downward as positive): $\sum F=mg-T=5a=49-T$.
3. Substitute $T=8a$ into the second equation: $49-8a=5a\to 13a=49\to a\approx 3.77{\text{ m/s}}^2$.
4. Solve for tension: $T=8\times 3.77\approx 30.2\text{ N}$.

Exam tip: If a pulley is accelerating (e.g., a movable pulley in a system), you must include forces on the pulley itself in your analysis; only fixed ideal pulleys have equal tension on both sides.

4. Combined Tension-Friction Connected Systems

Most AP Physics 1 problems involving both friction and tension are connected object systems, where one or more objects rest on a frictional surface, connected by a rope and pulley to a hanging object. The systematic approach to solve these problems is: 1) draw a separate free-body diagram for every object, 2) resolve forces into components aligned with the direction of possible motion, 3) write Newton’s second law for each object, using the equal tension and equal acceleration magnitude rules for ideal systems, 4) check if the system is stationary or accelerating by comparing the applied pulling force to maximum static friction, then solve the system of equations.

Worked Example

Block A (mass 4 kg) rests on a horizontal table, connected by an ideal rope over a fixed ideal pulley to hanging Block B (mass 3 kg). ${\mu }_s=0.35$ and ${\mu }_k=0.25$ between Block A and the table. Is the system stationary, or does it accelerate? If it accelerates, what is the tension?

1. Calculate maximum static friction on Block A: $N=m_{A}g=4\times 9.8=39.2\text{ N}$, so $f_{s,max}=0.35\times 39.2=13.72\text{ N}$.
2. Compare to the pulling force from Block B: For equilibrium, tension would need to equal $m_{B}g=29.4\text{ N}$. Since $29.4\text{ N}>13.72\text{ N}$, static friction cannot hold the system, so it accelerates.
3. Write Newton’s second law for the accelerating system: For Block A (right positive): $T-f_k=m_{A}a$, where $f_k=0.25\times 39.2=9.8\text{ N}$, so $T-9.8=4a$. For Block B (down positive): $29.4-T=3a$.
4. Add the equations to eliminate tension: $19.6=7a\to a=2.8{\text{ m/s}}^2$, so $T=4(2.8)+9.8=21\text{ N}$.

Exam tip: Always confirm the direction of friction: friction opposes impending or actual motion, so if the system is pulling a block up an incline, friction acts down the incline, and vice versa.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $f_s={\mu }_{s}N$ for static friction when the object is not at the point of sliding. Why: Students memorize the maximum static friction formula and apply it to all static friction cases, forgetting static friction adjusts to match the applied force. Correct move: Only use $f_s={\mu }_{s}N$ if the problem states the object is just about to slide; for all other stationary cases, use $f_s=F_{applied,parallel}$.
• Wrong move: Assuming normal force $N$ equals the object's weight $mg$. Why: Students generalize from simple horizontal surface problems to all cases. Correct move: Always calculate $N$ from Newton's second law in the direction perpendicular to the surface, accounting for angled applied forces or inclines before calculating friction.
• Wrong move: Assigning different acceleration magnitudes to connected objects on an ideal inextensible rope. Why: Students confuse different acceleration directions with different magnitudes. Correct move: For any two objects connected by an ideal rope, set the magnitude of acceleration equal when writing your system of equations.
• Wrong move: Changing the magnitude of tension when it goes around an ideal fixed pulley. Why: Students assume pulleys change tension, when they only change direction for ideal fixed pulleys. Correct move: For any ideal massless, frictionless pulley, tension has the same magnitude on both sides of the pulley.
• Wrong move: Using kinetic friction when the applied force is less than maximum static friction. Why: Students rush to use the kinetic friction formula without checking if motion occurs. Correct move: Always compare the net applied force trying to move the object to $f_{s,max}$ first; only use $f_k$ if the applied force exceeds $f_{s,max}$.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A 10 kg box rests on a horizontal surface with ${\mu }_s=0.5$ and ${\mu }_k=0.3$. A person pulls the box with a 30 N force at an angle of 30° above the horizontal. What is the magnitude of friction acting on the box? A) 0 N B) ~26 N C) ~36 N D) ~41 N

Worked Solution: First calculate the normal force. The vertical component of the applied force is $30\sin (30^{\circ })=15N$, so $N+15=mg=98N$, giving $N=83N$. Maximum static friction is $f_{s,max}=0.5\times 83=41.5N$. The horizontal component of the applied force is $30\cos (30^{\circ })\approx 26N$, which is less than $f_{s,max}$, so the box remains stationary. Static friction equals the horizontal applied force, so the magnitude is ~26 N. Correct answer: B

Question 2 (Free Response)

A block of mass $m_1=2\text{ kg}$ rests on an inclined plane that makes a 30° angle with the horizontal. The coefficient of static friction between the block and the plane is ${\mu }_s=0.3$, and ${\mu }_k=0.2$. The block is connected by an ideal rope running up the incline over a fixed ideal pulley to a hanging block of mass $m_2$. (a) Draw free-body diagrams for both blocks, labeling all forces. (b) What is the minimum mass $m_2$ required to hold the system stationary? (c) If $m_2=3\text{ kg}$, what is the magnitude of acceleration of the system, and what is the tension in the rope?

Worked Solution: (a) For $m_1$: Weight $m_{1}g$ downward, normal force $N$ perpendicular to the incline outward, tension $T$ up the incline, and static friction up the incline (opposing the impending downward slide of $m_1$ when $m_2$ is minimum). For $m_2$: Weight $m_{2}g$ downward, tension $T$ upward. (b) For stationary equilibrium, net force is zero on both objects. Perpendicular to the incline: $N=m_{1}g\cos \theta =2(9.8)(\cos 30^{\circ })\approx 17\text{ N}$. Maximum static friction $f_{s,max}={\mu }_{s}N=0.3(17)=5.1\text{ N}$. Along the incline: $T+f_{s,max}=m_{1}g\sin \theta =9.8\text{ N}$. For $m_2$, $T=m_{2}g$, so $m_{2}g=9.8-5.1=4.7$, giving $m_2\approx 0.48\text{ kg}$. (c) For $m_2=3\text{ kg}$, $m_1$ accelerates up the incline, friction acts down. Equations: $T-m_{1}g\sin \theta -{\mu }_{k}N=m_{1}a$ and $m_{2}g-T=m_{2}a$. Substitute values: $T-9.8-3.4=2a$ and $29.4-T=3a$. Add equations: $16.2=5a\to a=3.24{\text{ m/s}}^2$. Solve for tension: $T=29.4-3(3.24)\approx 19.7\text{ N}$.

Question 3 (Application / Real-World Style)

A moving company is lowering a 50 kg wooden safe down a 5 m long wooden ramp inclined at 20° to the horizontal, using a rope pulled by a worker parallel to the ramp to control speed. The coefficients of friction are ${\mu }_s=0.5$ and ${\mu }_k=0.3$. If the worker wants to lower the safe at constant speed, what is the tension the worker must apply? Is the tension an upward pull along the ramp or a downward pull?

Worked Solution: First calculate the weight components: parallel component $mg\sin \theta =50(9.8)(\sin 20^{\circ })\approx 167.6\text{ N}$, normal force $N=mg\cos \theta \approx 460\text{ N}$. The safe slides down, so kinetic friction acts upward along the ramp: $f_k={\mu }_{k}N=0.3(460)=138\text{ N}$. For constant speed, net force is zero: downward forces equal upward forces, so $167.6=T+138$, giving $T\approx 30\text{ N}$. In context: The worker must pull upward along the ramp with ~30 N of tension to keep the safe moving at constant speed, preventing it from accelerating down the ramp.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for all subsequent dynamics problems in AP Physics 1, and it will be immediately applied to circular motion and gravitation (Unit 3), where friction provides the centripetal force for objects like cars turning on flat roads. Tension also acts as the centripetal force for objects moving in vertical circles, so without mastering tension and friction analysis here, you will struggle to solve those centripetal force problems. Friction also appears later in energy problems, where it does non-conservative work, and in rotational dynamics when we analyze rolling motion without slipping, which relies entirely on static friction to provide torque. This topic also prepares you for more complex equilibrium problems that combine multiple contact forces.

• Dynamics: Newton's Laws
• Circular Motion: Centripetal Forces
• Work and Energy: Non-Conservative Work
• Rotational Motion: Rolling Without Slipping