Statics Quiz: Static And Kinetic Friction
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Static And Kinetic FrictionQuestion 1 of 9

A 40 N block is pulled across a horizontal surface by a rope making a 37° angle above the horizontal. The tension in the rope is 30 N. If the coefficient of kinetic friction is 0.25, what is the kinetic friction force?

5.5 N opposing the direction of motion
10.0 N opposing the direction of motion
7.5 N opposing the direction of motion
6.0 N opposing the direction of motion
4.5 N opposing the direction of motion
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Statics Quiz

Statics Quiz: Static And Kinetic Friction

Practice Static And Kinetic Friction in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Static And Kinetic Friction, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A 40 N block is pulled across a horizontal surface by a rope making a 37° angle above the horizontal. The tension in the rope is 30 N. If the coefficient of kinetic friction is 0.25, what is the kinetic friction force?

  1. 5.5 N opposing the direction of motion (correct answer)
  2. 10.0 N opposing the direction of motion
  3. 7.5 N opposing the direction of motion
  4. 6.0 N opposing the direction of motion
  5. 4.5 N opposing the direction of motion
Explanation: When analyzing friction problems with angled forces, you need to determine the normal force first, since kinetic friction depends on it through the relationship fk=μkNf_k = \mu_k N. Start by identifying all vertical forces acting on the block. The weight is 40 N downward. The rope tension of 30 N at 37° has a vertical component of 30sin⁡(37°)=30×0.6=18 N30 \sin(37°) = 30 \times 0.6 = 18 \text{ N} upward. For vertical equilibrium, the normal force plus the upward rope component must balance the weight: N+18=40N + 18 = 40, so N=22 NN = 22 \text{ N}. Now you can calculate the kinetic friction force: fk=μkN=0.25×22=5.5 Nf_k = \mu_k N = 0.25 \times 22 = 5.5 \text{ N}, which opposes the direction of motion. This confirms answer A is correct. Let's examine why the other answers are wrong. Answer B (10.0 N) would result from incorrectly using fk=0.25×40=10 Nf_k = 0.25 \times 40 = 10 \text{ N}, which ignores the rope's vertical component that reduces the normal force. Answer C (7.5 N) might come from using the horizontal component of tension (30cos⁡(37°)=24 N30 \cos(37°) = 24 \text{ N}) in some incorrect calculation. Answer D (6.0 N) could result from various computational errors or using an incorrect normal force value. The key strategy here is always finding the normal force first when dealing with angled applied forces. The vertical component of any angled force will affect the normal force, which directly impacts the friction calculation. Don't assume the normal force equals the object's weight when other vertical forces are present.

Question 2

A ladder of weight 200 N leans against a frictionless wall at 60° from the horizontal. The coefficient of static friction between the ladder and ground is 0.6. What is the friction force at the ground when the ladder is in equilibrium?

  1. 57.7 N directed horizontally away from the wall (correct answer)
  2. 100.0 N directed horizontally away from the wall
  3. 115.5 N directed horizontally away from the wall
  4. 86.6 N directed horizontally away from the wall
  5. 173.2 N directed horizontally away from the wall
Explanation: When analyzing a ladder leaning against a wall, you're dealing with a classic static equilibrium problem that requires both force and moment balance. The key insight is that the wall is frictionless, so it can only provide a normal force perpendicular to its surface. Let's set up the equilibrium equations. The forces acting on the ladder are: its weight (200 N downward at the center), the normal force from the wall (NwN_w horizontal), the normal force from the ground (NgN_g vertical), and friction from the ground (ff horizontal). For horizontal force equilibrium: f=Nwf = N_w For moment equilibrium about the base of the ladder, if the ladder length is LL: The weight creates a clockwise moment of 200×L2×cos⁡(60°)=50L200 \times \frac{L}{2} \times \cos(60°) = 50L, and the wall's normal force creates a counterclockwise moment of Nw×L×sin⁡(60°)=Nw×L×32N_w \times L \times \sin(60°) = N_w \times L \times \frac{\sqrt{3}}{2}. Setting these equal: 50L=Nw×L×3250L = N_w \times L \times \frac{\sqrt{3}}{2} Solving: Nw=1003=57.7 NN_w = \frac{100}{\sqrt{3}} = 57.7 \text{ N} Therefore, the friction force equals 57.7 N directed away from the wall. Answer A is correct at 57.7 N. Answer B (100.0 N) incorrectly uses the full weight without considering geometry. Answer C (115.5 N) appears to double the correct value. Answer D (86.6 N) might result from using sine instead of cosine in the moment calculation. Remember: Always draw a free body diagram and use both force equilibrium and moment equilibrium equations. The coefficient of friction given is often a red herring—check if the actual friction needed exceeds the maximum available.

Question 3

A car traveling at 20 m/s begins braking on a horizontal road. If the coefficient of kinetic friction between tires and road is 0.7, what is the magnitude of the friction force per unit mass during braking?

  1. 6.86 N/kg acting opposite to the direction of motion (correct answer)
  2. 14.0 N/kg acting opposite to the direction of motion
  3. 9.8 N/kg acting opposite to the direction of motion
  4. 20.0 N/kg acting opposite to the direction of motion
  5. 2.86 N/kg acting opposite to the direction of motion
Explanation: When you encounter friction problems in statics, remember that kinetic friction force depends on the normal force and coefficient of friction. Here, you need to find the friction force per unit mass during braking. The friction force is given by fk=μkNf_k = \mu_k N, where μk=0.7\mu_k = 0.7 and NN is the normal force. On a horizontal surface, the normal force equals the weight: N=mgN = mg. Therefore, fk=μkmg=0.7mgf_k = \mu_k mg = 0.7mg. To find the friction force per unit mass, divide by mass: fkm=0.7mgm=0.7g=0.7×9.8=6.86 N/kg\frac{f_k}{m} = \frac{0.7mg}{m} = 0.7g = 0.7 \times 9.8 = 6.86 \text{ N/kg}. Since friction opposes motion, it acts opposite to the car's direction of travel. Looking at the wrong answers: Choice B (14.0 N/kg) appears to use μk×2g\mu_k \times 2g, possibly from incorrectly doubling the gravitational acceleration. Choice C (9.8 N/kg) uses just gg without applying the coefficient of friction—this would be the friction force if μk=1\mu_k = 1. Choice D (20.0 N/kg) mistakenly uses the initial velocity (20 m/s) as the force per unit mass, confusing speed with force. Study tip: In friction problems, always remember that kinetic friction force equals μk\mu_k times the normal force. On horizontal surfaces, the normal force simply equals the weight (mgmg). The initial velocity doesn't directly determine the friction force—only the coefficient of friction and normal force matter.

Question 4

A 30 kg box is pushed up a 20° ramp with constant velocity. The pushing force is parallel to the ramp surface. If the coefficient of kinetic friction is 0.25, what is the required pushing force?

  1. 169.4 N applied parallel to the ramp surface upward (correct answer)
  2. 100.4 N applied parallel to the ramp surface upward
  3. 138.6 N applied parallel to the ramp surface upward
  4. 276.2 N applied parallel to the ramp surface upward
  5. 207.3 N applied parallel to the ramp surface upward
Explanation: When analyzing forces on an inclined plane with constant velocity, you're dealing with equilibrium conditions where all forces balance. Since the box moves at constant velocity, the net force is zero, meaning the pushing force must exactly counteract both the component of weight down the ramp and the friction force. Start by breaking the weight into components. The component parallel to the ramp (down the slope) is mgsin⁡(20°)=30×9.8×sin⁡(20°)=100.4 Nmg\sin(20°) = 30 × 9.8 × \sin(20°) = 100.4 \text{ N}. The component perpendicular to the ramp is mgcos⁡(20°)=30×9.8×cos⁡(20°)=276.2 Nmg\cos(20°) = 30 × 9.8 × \cos(20°) = 276.2 \text{ N}, which equals the normal force. The kinetic friction force opposes motion up the ramp: fk=μkN=0.25×276.2=69.0 Nf_k = μ_k N = 0.25 × 276.2 = 69.0 \text{ N} For equilibrium parallel to the ramp: Fpush=mgsin⁡(20°)+fk=100.4+69.0=169.4 NF_{push} = mg\sin(20°) + f_k = 100.4 + 69.0 = 169.4 \text{ N} Answer A (169.4 N) correctly accounts for both weight component and friction. Answer B (100.4 N) only considers the weight component down the ramp while ignoring friction entirely. Answer C (138.6 N) likely results from an error in the friction calculation or angle usage. Answer D (276.2 N) mistakenly uses the normal force value instead of properly calculating the required pushing force. Remember: on inclined plane problems with friction, you must always account for both the gravitational component along the plane AND the friction force. Missing either component will lead to an incorrect result.

Question 5

A 50 N block rests on a horizontal surface with a coefficient of static friction μs=0.4\mu_s = 0.4 and coefficient of kinetic friction μk=0.3\mu_k = 0.3. A horizontal force FF is gradually increased from zero. What is the friction force when F=15F = 15 N?

  1. 15 N opposing the applied force (correct answer)
  2. 20 N opposing the applied force
  3. 15 N in the direction of the applied force
  4. 0 N because the block is not moving
  5. 35 N opposing the applied force
Explanation: When analyzing friction problems, you need to understand that static friction is a responsive force that adjusts to match the applied force, up to its maximum limit. The block will remain stationary as long as the applied force doesn't exceed the maximum static friction force. First, calculate the maximum static friction: fs,max=μs×N=0.4×50 N=20 Nf_{s,max} = \mu_s \times N = 0.4 \times 50\text{ N} = 20\text{ N}. Since the applied force F=15 NF = 15\text{ N} is less than this maximum, the block remains stationary. For a stationary object, static friction equals the applied force to maintain equilibrium. Therefore, the friction force is exactly 15 N, directed opposite to the applied force to balance it out. This confirms answer A is correct. B is wrong because 20 N represents the maximum possible static friction, not the actual friction when F=15 NF = 15\text{ N}. Static friction only reaches its maximum value at the instant before sliding begins. C is incorrect because friction always opposes motion or potential motion. If friction acted in the same direction as the applied force, it would accelerate the block rather than maintain equilibrium. D reflects a common misconception that stationary objects experience no friction. While kinetic friction requires motion, static friction exists whenever there's a tendency to move, even if no actual motion occurs. Study tip: Remember that static friction is a "lazy" force—it only provides exactly what's needed to prevent motion, up to its maximum limit. Always compare the applied force to the maximum static friction to determine if the object will move.

Question 6

A 25 kg crate slides down a ramp inclined at 30° with kinetic friction coefficient μk=0.2\mu_k = 0.2. What is the magnitude of the kinetic friction force acting on the crate?

  1. 42.4 N directed up the ramp parallel to the surface (correct answer)
  2. 49.0 N directed up the ramp parallel to the surface
  3. 122.5 N directed up the ramp parallel to the surface
  4. 42.4 N directed perpendicular to the ramp surface
  5. 24.5 N directed up the ramp parallel to the surface
Explanation: When analyzing friction on inclined planes, you need to understand that kinetic friction always opposes motion and depends on the normal force, not the weight directly. To find the kinetic friction force, use fk=μkNf_k = \mu_k N, where NN is the normal force. On an inclined plane, the normal force equals the component of weight perpendicular to the surface: N=mgcos⁡θN = mg\cos\theta. Here, N=(25 kg)(9.8 m/s2)cos⁡(30°)=245×0.866=212.2 NN = (25\text{ kg})(9.8\text{ m/s}^2)\cos(30°) = 245 × 0.866 = 212.2\text{ N}. Therefore, fk=0.2×212.2=42.4 Nf_k = 0.2 × 212.2 = 42.4\text{ N}. Since the crate slides down the ramp, friction acts up the ramp parallel to the surface to oppose this motion. Looking at the wrong answers: Answer B (49.0 N) incorrectly uses mgsin⁡θmg\sin\theta instead of mgcos⁡θmg\cos\theta for the normal force calculation—a common mistake where students confuse the parallel and perpendicular components. Answer C (122.5 N) appears to use the full weight (245 N) multiplied by 0.5, suggesting confusion about which force component to use. Answer D has the correct magnitude (42.4 N) but incorrectly states the direction as perpendicular to the ramp—friction forces always act parallel to contact surfaces, never perpendicular. Remember this key pattern: on inclined planes, always break weight into components first. Normal force uses the cosine component (perpendicular to surface), and kinetic friction always opposes the direction of motion along the surface.

Question 7

A car travels around a horizontal circular track of radius 50 m. The coefficient of static friction between the tires and road is μs=0.7\mu_s = 0.7. If the car's mass is 1200 kg, what happens when the car attempts to maintain a constant speed of 25 m/s?

  1. The car maintains circular motion with friction providing 15,000 N centripetal force
  2. The car slides outward because required centripetal force exceeds available friction (correct answer)
  3. The car automatically reduces speed to 18.7 m/s to match friction limit
  4. The car maintains motion with tire deformation reducing effective radius
Explanation: For circular motion, the centripetal acceleration required is ac=v2/r=(25)2/50=12.5a_c = v^2/r = (25)^2/50 = 12.5 m/s². The centripetal force required is Fc=mac=1200×12.5=15,000F_c = ma_c = 1200 \times 12.5 = 15,000 N. This force must be provided by friction between tires and road. The maximum friction available is fmax=μsmg=0.7×1200×9.81=8,237f_{max} = \mu_s mg = 0.7 \times 1200 \times 9.81 = 8,237 N. Since the required centripetal force (15,000 N) exceeds the maximum available friction (8,237 N), the car cannot maintain this circular path and will begin to slide outward. The maximum safe speed would be vmax=μsgr=0.7×9.81×50=18.5v_{max} = \sqrt{\mu_s gr} = \sqrt{0.7 \times 9.81 \times 50} = 18.5 m/s.

Question 8

A uniform sphere of radius R and mass M rolls without slipping down an inclined plane of angle θ. The coefficient of static friction is μs\mu_s. What is the minimum coefficient of static friction required to ensure rolling without slipping?

  1. μs=27tan⁡θ\mu_s = \frac{2}{7}\tan\theta based on rotational dynamics of the sphere (correct answer)
  2. μs=13tan⁡θ\mu_s = \frac{1}{3}\tan\theta considering translational motion constraints only
  3. μs=25tan⁡θ\mu_s = \frac{2}{5}\tan\theta accounting for moment of inertia effects
  4. μs=tan⁡θ\mu_s = \tan\theta since friction must balance the parallel component of weight
Explanation: For a sphere rolling without slipping, we have the constraint a=αRa = \alpha R where a is linear acceleration and α is angular acceleration. The forces acting are: weight component down the incline (Mgsin⁡θMg\sin\theta), normal force (Mgcos⁡θMg\cos\theta), and friction up the incline (f). Newton's second law: Mgsin⁡θ−f=MaMg\sin\theta - f = Ma. For rotation about center of mass: fR=Iα=25MR2⋅aR=25MRafR = I\alpha = \frac{2}{5}MR^2 \cdot \frac{a}{R} = \frac{2}{5}MRa. Therefore: f=25Maf = \frac{2}{5}Ma. Substituting into the first equation: Mgsin⁡θ−25Ma=MaMg\sin\theta - \frac{2}{5}Ma = Ma, which gives a=5gsin⁡θ7a = \frac{5g\sin\theta}{7}. The required friction is f=25M⋅5gsin⁡θ7=2Mgsin⁡θ7f = \frac{2}{5}M \cdot \frac{5g\sin\theta}{7} = \frac{2Mg\sin\theta}{7}. For no slipping: f≤μsN=μsMgcos⁡θf \leq \mu_s N = \mu_s Mg\cos\theta. Therefore: 2Mgsin⁡θ7≤μsMgcos⁡θ\frac{2Mg\sin\theta}{7} \leq \mu_s Mg\cos\theta, giving μs≥2tan⁡θ7\mu_s \geq \frac{2\tan\theta}{7}.

Question 9

A uniform rod of length L and mass M is placed on two supports: one at distance L/4 from the left end, and another at distance L/4 from the right end. The coefficient of static friction at both supports is μs\mu_s. What is the minimum value of μs\mu_s required to prevent the rod from slipping when it is in equilibrium?

  1. μs=13\mu_s = \frac{1}{3} based on moment equilibrium about either support point
  2. μs=12\mu_s = \frac{1}{2} considering the rod's tendency to slide horizontally
  3. μs=23\mu_s = \frac{2}{3} accounting for maximum moment arm effects
  4. No friction required since the rod is symmetric and in stable equilibrium (correct answer)
Explanation: This is a trick question that tests understanding of when friction is actually required. The rod is uniform with mass M, so its center of gravity is at L/2 from either end. The supports are at L/4 and 3L/4 from the left end, making them symmetric about the center of gravity. Since the rod is symmetric and the supports are equidistant from the center of mass, each support carries exactly Mg/2 vertically upward. There are no horizontal forces applied to the system, and the weight acts vertically downward through the center. Since there's no tendency for horizontal motion and the vertical forces are balanced, no friction force is required at either support. The rod will remain in stable equilibrium without any friction. The other answers incorrectly assume there must be some slipping tendency requiring friction.