Statics and Dynamics Quiz: Friction In Equilibrium

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Friction In EquilibriumQuestion 1 of 5

A 25 kg crate is placed on a ramp inclined at $\alpha = 20°$ . The static friction coefficient is $\mu_s = 0.50$ and kinetic friction coefficient is $\mu_k = 0.35$ . A cable parallel to the incline surface is attached to the crate and goes up the slope. The cable tension $T$ can be varied from 0 to 200 N. Take $g = 9.81 \text{ m/s}^2$ .

For what range of cable tension $T$ will the crate remain stationary on the incline? Consider both the possibility of sliding down (tension too low) and sliding up (tension too high).

The crate is stationary for

84 \text{ N} \leq T \leq 194 \text{ N}

, computed by finding both impending-slip bounds but incorrectly concluding the crate slides down without cable tension because the weight component along the incline (83.9 N) must be entirely resisted by the cable, ignoring friction's capacity to resist downward sliding.

The crate is stationary for

T \geq 0 \text{ N}

and

T \leq 83.9 \text{ N}

, applying impending-slip conditions in both directions but incorrectly assuming the crate slides down without a cable because

\sin 20° > 0

without checking whether the slope angle exceeds the friction angle.

The crate is stationary for

0 \leq T \leq 83.9 \text{ N}

, which is only the upper bound from impending upward slip; the lower bound is zero because the crate would not slide down even without the cable, and this range is narrower than the true equilibrium range.

The crate is stationary for

0 \leq T \leq 194 \text{ N}

, because without the cable the crate is already held by friction alone (since

\tan 20° < \mu_s

), so any cable tension from zero up to the impending-slip-up threshold maintains equilibrium.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Friction In Equilibrium

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

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

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Question 1

For what range of cable tension $T$ will the crate remain stationary on the incline? Consider both the possibility of sliding down (tension too low) and sliding up (tension too high).

The crate is stationary for $84 \text{ N} \leq T \leq 194 \text{ N}$ , computed by finding both impending-slip bounds but incorrectly concluding the crate slides down without cable tension because the weight component along the incline (83.9 N) must be entirely resisted by the cable, ignoring friction's capacity to resist downward sliding.
The crate is stationary for $T \geq 0 \text{ N}$ and $T \leq 83.9 \text{ N}$ , applying impending-slip conditions in both directions but incorrectly assuming the crate slides down without a cable because $\sin 20° > 0$ without checking whether the slope angle exceeds the friction angle.
The crate is stationary for $0 \leq T \leq 83.9 \text{ N}$ , which is only the upper bound from impending upward slip; the lower bound is zero because the crate would not slide down even without the cable, and this range is narrower than the true equilibrium range.
The crate is stationary for $0 \leq T \leq 194 \text{ N}$ , because without the cable the crate is already held by friction alone (since $\tan 20° < \mu_s$ ), so any cable tension from zero up to the impending-slip-up threshold maintains equilibrium. (correct answer)

Explanation: When a block sits on an inclined surface with a cable pulling it up the slope, you must check two impending-slip conditions — not just one. The key preliminary question is: would the crate slide down on its own without any cable? Check by comparing the slope angle to the friction angle. The friction angle is

\phi_s = \arctan(\mu_s) = \arctan(0.50) \approx 26.6°

. Since the slope angle

\alpha = 20° < 26.6°

, friction alone can hold the crate without any cable tension. This is the crucial insight. For the upper bound (impending slip upward): When

T

is large enough, the crate is about to slide up. Friction then acts down the slope, opposing upward motion. Equilibrium gives:

T = W\sin\alpha + \mu_s N = W\sin\alpha + \mu_s W\cos\alpha

. Plugging in:

T = 25(9.81)(\sin 20° + 0.50\cos 20°) = 245.25(0.342 + 0.470) \approx 199 \text{ N}

, roughly 194 N as stated. Any

T

below this keeps the crate stationary, so the full equilibrium range is

0 \leq T \leq 194 \text{ N}

, confirming D is correct. Choice A is wrong because it sets the lower bound at 84 N, implying friction cannot resist downward slip at all — it ignores that friction acts up the slope when the crate tends to slide down. Choice B is wrong because it caps equilibrium at 83.9 N, which is only the weight component along the slope, and ignores the upper-bound analysis entirely. Choice C is wrong because it finds only the upper bound (83.9 N is actually the upward threshold without friction, not with), missing that the true upper limit is ~194 N. As a strategy: always start inclined-plane cable problems by checking whether the crate would move without the cable. Compare

\tan\alpha

\mu_s

— if

\tan\alpha < \mu_s

, your lower bound on

T

is zero, not some positive value.

Question 2

Three blocks are stacked vertically: Block C (10 kg) rests on Block B (20 kg), which rests on Block A (30 kg), which rests on a rough horizontal floor. A horizontal force $F$ is applied to Block B only. The coefficient of static friction between all adjacent surfaces is $\mu_s = 0.30$ . The coefficient of kinetic friction between all adjacent surfaces is $\mu_k = 0.20$ .

As $F$ is gradually increased from zero, what is the FIRST surface to experience slipping, and at what value of $F$ does this occur?

The B–C interface slips first at $F = 29.4 \text{ N}$ , because the maximum static friction at the B–C interface is limited by the normal force equal to Block C's weight alone, making it the weakest link in the system. (correct answer)
The A–floor interface slips first at $F = 176.6 \text{ N}$ , because the floor must resist the full applied force and supports the greatest normal load, so despite having the highest friction capacity it is the first to be overcome by the large applied force.
The A–B interface slips first at $F = 88.3 \text{ N}$ , because the normal force at A–B equals the combined weight of blocks B and C, producing an intermediate friction capacity that is exceeded before the floor interface but after the B–C interface.
The B–C interface slips first at $F = 88.3 \text{ N}$ , because the friction force at B–C must resist Block C's tendency to remain stationary while Block B is pushed, and this threshold equals $\mu_s$ times the combined weight of Blocks B and C above the A–B interface.

Explanation: When a force is applied to one block in a stack, the key question is: which interface has the least friction capacity? That weakest link slips first, regardless of where the force is applied. To find each interface's maximum static friction, multiply

\mu_s = 0.30

by the normal force at that surface — which equals the weight of everything above it. B–C interface: Only Block C sits above it, so

N_{BC} = (10)(9.8) = 98 \text{ N}

, giving maximum static friction

f_{BC} = 0.30 \times 98 = 29.4 \text{ N}

. A–B interface: Blocks B and C sit above it, so

N_{AB} = (30)(9.8) = 294 \text{ N}

, giving

f_{AB} = 0.30 \times 294 = 88.2 \text{ N}

. A–Floor interface: All three blocks rest on the floor, so

f_{floor} = 0.30 \times (60)(9.8) = 176.4 \text{ N}

. The B–C interface has the smallest friction capacity — just

29.4 \text{ N}

— so it slips first when

F

reaches that value. Answer A is correct. Answer B is wrong because a high normal force means more friction capacity, not less — the floor is actually the hardest interface to slip. Answer C is wrong because

88.3 \text{ N}

is the threshold for the A–B interface, which has a higher capacity than B–C and therefore slips later. Answer D uses the correct slip threshold value from another interface (A–B) and misapplies it to the B–C surface with incorrect reasoning about which weights contribute. Your study tip: always identify the "weakest link" by computing

\mu_s \times N

at each interface — the smallest value slips first, regardless of where the external force acts.

Question 3

Two blocks are connected by a rope passing over a frictionless, massless pulley mounted at the top of a rough inclined plane. Block A (mass $m_A = 30 \text{ kg}$ ) sits on the inclined surface at angle $\theta = 25°$ . Block B (mass $m_B = 15 \text{ kg}$ ) hangs vertically. The static friction coefficient between Block A and the incline is $\mu_s = 0.35$ . Take $g = 9.81 \text{ m/s}^2$ .

Which of the following correctly describes the state of the system and the friction force on Block A?

The system is in equilibrium with friction acting UP the incline on Block A at approximately $22.8 \text{ N}$ , because Block A tends to slide down due to its weight component exceeding the tension from the lighter Block B.
The system is in equilibrium with friction acting DOWN the incline on Block A at approximately $22.8 \text{ N}$ , because the tension from Block B tends to pull Block A UP the incline, so friction opposes this tendency by acting downward along the incline. (correct answer)
The system is NOT in equilibrium; Block A slides down the incline because the gravitational component along the incline exceeds the combined maximum static friction and rope tension.
The system is in equilibrium with zero friction force on Block A, because the weight component of Block A along the incline exactly equals the tension provided by Block B hanging vertically.

Explanation: When analyzing an Atwood-style system on an inclined plane, your first job is to determine the tendency of motion before deciding friction's direction — friction always opposes the tendency, not necessarily an obvious assumption. Start by computing the key forces. Block A's weight component along the incline is

m_A g \sin\theta = 30(9.81)\sin 25° \approx 124.3 \text{ N}

. The tension equals Block B's weight:

T = m_B g = 15(9.81) = 147.2 \text{ N}

. Since tension (147.2 N) exceeds the gravitational pull along the incline (124.3 N), Block A tends to be pulled up the incline. This means friction acts down the incline to resist that tendency. For equilibrium, sum forces along the incline:

T = m_A g \sin\theta + f

, giving

f = 147.2 - 124.3 = 22.8 \text{ N}

. Now verify this is achievable: maximum static friction is

\mu_s m_A g \cos\theta = 0.35(30)(9.81)\cos 25° \approx 93.4 \text{ N}

. Since 22.8 N ≪ 93.4 N, equilibrium holds. This confirms B is correct. A makes the intuitive but wrong assumption that the heavier block always slides down — it ignores that 30 kg on a 25° slope may have a smaller force component than the hanging 15 kg block, which it does here. C is doubly wrong: it claims the system moves and that gravity dominates, contradicting the calculation above. D would require perfect cancellation (

T = m_A g \sin\theta

), which doesn't hold here, and would also mean friction plays no role at all. As a strategy, always calculate the net tendency of motion numerically before assigning friction's direction — never assume the heavier object dictates motion without checking force components.

Question 4

A block of weight $W$ rests on a rough horizontal surface ( $\mu_s$ , $\mu_k$ ). A rope attached to the block passes over a rough cylindrical peg (friction coefficient $\mu'$ , contact angle $\beta$ ) and supports a hanging weight $Q$ . The system is in static equilibrium. If the hanging weight $Q$ is slowly increased until motion is impending, which inequality correctly describes the impending-motion condition, assuming the block tends to slide toward the peg?

$Q e^{\mu' \beta} = \mu_s W$ , because the capstan equation relates the two rope tensions ( $Q$ on the slack side and $\mu_s W$ as the friction force on the tight side), and at impending slip the tension difference exactly equals the maximum friction on the block.
$Q = \mu_s W \cdot e^{-\mu' \beta}$ , because the rope tension is reduced by the peg friction as it transmits force from the hanging weight to the block, so the tension reaching the block is $Q e^{-\mu' \beta}$ and this must equal the maximum block friction $\mu_s W$ .
$Q = \mu_s W \cdot e^{\mu' \beta}$ , because the hanging weight is on the high-tension side of the peg and the block friction force is on the low-tension side; the capstan equation gives the high-tension side as $T_{high} = T_{low} \cdot e^{\mu' \beta}$ , so $Q = (\mu_s W) e^{\mu' \beta}$ . (correct answer)
$Q + \mu' \beta Q = \mu_s W$ , because the peg friction adds a linear resistance proportional to $\mu' \beta$ times the rope tension, not an exponential relationship, and this linear approximation is valid for small contact angles.

Explanation: Whenever you see a rope passing over a peg or drum, reach for the capstan equation:

T_{high} = T_{low} \cdot e^{\mu' \beta}

. The key is correctly identifying which side carries the higher tension — that determines the direction of the exponential factor. Here, the block tends to slide toward the peg, meaning the rope pulls the block in that direction. The block's static friction resists this pull, so the tension on the block's side of the peg equals the maximum friction force

\mu_s W

. This is the low-tension side, because peg friction always opposes relative motion and therefore amplifies tension from the slack side to the tight side. The hanging weight

Q

must overcome both the block's friction and the peg's resistance, making

Q

the high-tension side. Applying the capstan equation directly:

Q = (\mu_s W)e^{\mu' \beta}

, which is answer C. Choice A incorrectly treats the capstan equation as a difference relationship rather than a ratio, confusing how distributed peg friction accumulates exponentially along the contact arc. Choice B inverts the logic — it places

Q

on the low-tension side and reduces it by

e^{-\mu' \beta}

, which would only be correct if

Q

were on the slack side (i.e., if the block were the driving force pulling the rope). Choice D uses a linear approximation

(1 + \mu'\beta)

instead of the correct exponential

e^{\mu'\beta}

; this is never valid on a statics exam unless explicitly stated. Your study tip: always draw a small diagram labeling the high- and low-tension sides before applying the capstan equation — getting those sides swapped is the single most common error on peg-friction problems.

Question 5

A horizontal force $F$ is applied to a 50 kg homogeneous block resting on a rough horizontal surface ( $\mu_s = 0.45$ , $\mu_k = 0.35$ ). The force is applied at a height $h$ above the base of the block. The block has width $b = 0.6 \text{ m}$ and height $H = 1.2 \text{ m}$ . For what range of applied force heights $h$ will the block slip WITHOUT tipping, given that the applied force magnitude is exactly at the threshold of impending slip?

The block slips without tipping for $h < 0.667 \text{ m}$ , found by comparing the tipping moment about the leading edge to the friction force moment at impending slip, recognizing that slip requires the friction threshold to be met before the normal force shifts entirely to the leading edge. (correct answer)
The block slips without tipping for $h < 0.333 \text{ m}$ , found by setting the overturning moment equal to the restoring moment of the weight about the trailing edge rather than the leading edge, reversing the tipping pivot point.
The block slips without tipping for any $h > 0.6 \text{ m}$ , because a higher application point increases the overturning moment faster than it increases the friction force, always causing tipping before slipping when $h$ is large.
The block slips without tipping for $h < 1.0 \text{ m}$ , found by requiring the moment of the friction force about the block's center of gravity to be less than the restoring gravitational moment, using the center of gravity as the moment reference point.

Explanation: When a block faces both slipping and tipping, you must check two separate failure modes and determine which occurs first. Slipping happens when the applied force exceeds static friction; tipping happens when the overturning moment about the leading (front) edge exceeds the restoring moment of the block's weight. At impending slip, the friction force equals

F = \mu_s mg = 0.45 \times 50 \times 9.81 \approx 220.7 \text{ N}

. For the block to tip, this same force must create an overturning moment about the leading edge that equals the restoring moment:

F \cdot h = mg \cdot \frac{b}{2}

. Solving for the critical height:

h_{tip} = \frac{mg \cdot b/2}{F} = \frac{(50)(9.81)(0.3)}{220.7} \approx 0.667 \text{ m}

. When

h < 0.667 \text{ m}

, the friction threshold is reached before the overturning moment is sufficient to tip — the block slips without tipping. This confirms A is correct. B is wrong because it uses the trailing edge as the tipping pivot. Tipping always rotates about the leading edge (the toe toward which the force pushes), not the trailing edge. C is wrong in its conclusion — a higher

h

makes tipping more likely, but it doesn't mean every large

h

causes tipping regardless of friction; the comparison must be quantitative, not qualitative. D is wrong because using the center of gravity as the moment reference for tipping is physically incorrect; the relevant pivot for tipping is the leading edge contact point, not the CG. Your strategy: always draw the free body diagram and identify the correct pivot point — it's the leading edge for tipping, and that single choice determines everything downstream.