Statics and Dynamics Quiz: Dry Friction Model

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Dry Friction ModelQuestion 1 of 6

A 30 kg crate is pushed across a rough horizontal floor by a force $P$ applied at an angle $\theta$ below the horizontal (i.e., the force has a downward component). The kinetic coefficient of friction is $\mu_k = 0.40$ and the static coefficient is $\mu_s = 0.55$ . The crate moves at constant velocity.

As the push angle $\theta$ is increased from 0° toward 90° (keeping the crate moving at constant velocity by adjusting $|P|$ accordingly), what happens to the required magnitude of $P$ ?

P

first decreases then increases without bound, because increasing

\theta

initially reduces the horizontal component needed to overcome friction, but eventually the increased normal force (and thus friction) dominates and

P

rises steeply, becoming infinite at a critical angle where no finite

P

can sustain motion.

P

monotonically increases and eventually becomes infinite at a critical angle

\theta^* = \arctan(1/\mu_k) \approx 68.2°

, because the denominator in

P = \mu_k mg/(\cos\theta - \mu_k\sin\theta)

decreases continuously from 1 toward zero as

\theta

increases, making

P

grow without bound.

P

monotonically decreases as

\theta

increases, because the horizontal component

P\cos\theta

must equal

\mu_k N

, and a larger

\theta

allows the same friction force to be overcome with a smaller

P

due to the geometric leverage of the inclined push.

P

remains constant as

\theta

increases, because the equilibrium condition

P\cos\theta = \mu_k(mg + P\sin\theta)

scales proportionally in

P

, leaving the required magnitude unchanged as long as motion is maintained at constant velocity.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Dry Friction Model

Practice Dry Friction Model in Statics and Dynamics 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 Dry Friction Model, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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.

All questions

Question 1

As the push angle $\theta$ is increased from 0° toward 90° (keeping the crate moving at constant velocity by adjusting $|P|$ accordingly), what happens to the required magnitude of $P$ ?

$P$ first decreases then increases without bound, because increasing $\theta$ initially reduces the horizontal component needed to overcome friction, but eventually the increased normal force (and thus friction) dominates and $P$ rises steeply, becoming infinite at a critical angle where no finite $P$ can sustain motion.
$P$ monotonically increases and eventually becomes infinite at a critical angle $\theta^* = \arctan(1/\mu_k) \approx 68.2°$ , because the denominator in $P = \mu_k mg/(\cos\theta - \mu_k\sin\theta)$ decreases continuously from 1 toward zero as $\theta$ increases, making $P$ grow without bound. (correct answer)
$P$ monotonically decreases as $\theta$ increases, because the horizontal component $P\cos\theta$ must equal $\mu_k N$ , and a larger $\theta$ allows the same friction force to be overcome with a smaller $P$ due to the geometric leverage of the inclined push.
$P$ remains constant as $\theta$ increases, because the equilibrium condition $P\cos\theta = \mu_k(mg + P\sin\theta)$ scales proportionally in $P$ , leaving the required magnitude unchanged as long as motion is maintained at constant velocity.

Explanation: When a pushing force is applied below the horizontal, both friction and the normal force depend on

P

. Setting up equilibrium (constant velocity means zero net force) gives you two equations: horizontally,

P\cos\theta = \mu_k N

, and vertically,

N = mg + P\sin\theta

. Substituting the second into the first and solving for

P

yields the key formula:

P = \frac{\mu_k mg}{\cos\theta - \mu_k \sin\theta}

Notice the denominator. As

\theta

increases from 0°,

\cos\theta

shrinks while

\mu_k\sin\theta

grows — so the denominator continuously decreases toward zero. When it hits zero,

P \to \infty

. That critical angle occurs at

\cos\theta^* = \mu_k\sin\theta^*

, or

\theta^* = \arctan(1/\mu_k) = \arctan(2.5) \approx 68.2°

. This confirms B:

P

grows monotonically and becomes infinite at that critical angle — beyond it, no finite push can sustain motion because the downward component keeps increasing friction faster than the horizontal component can overcome it. A is tempting but wrong — it describes a scenario where

P

first drops, implying an optimal angle exists. That would be true if the force were applied above the horizontal (pulling), where a rising

\theta

initially reduces normal force and friction. Pushing always worsens friction as

\theta

increases, so there is no minimum. C incorrectly claims

P

decreases, misreading the geometric effect — the shrinking

\cos\theta

in the denominator outpaces any benefit. D claims

P

is constant, ignoring that

\theta

fundamentally changes the denominator. Your study tip: always check whether a force is pushing into or pulling away from the surface — this flips whether friction gets better or worse with angle, and it's a classic trap on statics problems.

Question 2

A block of mass $m$ rests on a rough inclined plane of angle $\alpha$ . The static friction coefficient is $\mu_s$ and the kinetic friction coefficient is $\mu_k < \mu_s$ . The block is given a brief upward push along the incline and then released.

After the push is removed, which condition on $\mu_s$ and $\alpha$ correctly determines whether the block comes to rest and stays at rest (rather than sliding back down after stopping)?

$\mu_k \geq \tan\alpha$ , because once the block is moving downward after stopping, the kinetic friction force must be sufficient to halt it; the static coefficient is irrelevant once the block has begun to slide back.
$\mu_s \geq \tan\alpha$ , because after the block decelerates and stops, the static friction force required to hold it equals $mg\sin\alpha$ , and the maximum available static friction is $\mu_s mg\cos\alpha$ ; the block stays put if and only if this maximum meets or exceeds the gravitational component. (correct answer)
$\mu_k \geq \tan\alpha$ , because the block decelerates due to both gravity and kinetic friction acting downward (during upward travel both oppose motion), and the same kinetic friction must now reverse to prevent sliding back, so $\mu_k$ — not $\mu_s$ — is the operative coefficient at the moment of stopping.
Both $\mu_s \geq \tan\alpha$ and $\mu_k \geq \tan\alpha$ must hold simultaneously, because $\mu_s$ governs initial resistance to sliding back while $\mu_k$ determines the deceleration rate during upward travel; if either condition fails, the block will not remain stationary after stopping.

Explanation: Whenever a dynamics question asks about a block "coming to rest and staying at rest," you need to recognize that two separate physical phases are involved: the moving phase (governed by kinetic friction) and the stationary phase (governed by static friction). Conflating these two phases is exactly the trap this question sets. After the push is removed, the block travels upward while both gravity and kinetic friction act downward along the incline — it decelerates and eventually stops. At the instant it stops, kinetic friction vanishes entirely. The question then becomes: can static friction hold the block in place? The gravitational component pulling the block down the incline is

mg\sin\alpha

, and the maximum static friction force available is

\mu_s mg\cos\alpha

. The block stays put if and only if

\mu_s mg\cos\alpha \geq mg\sin\alpha

, which simplifies to

\mu_s \geq \tan\alpha

. This is precisely answer B, and it is the correct answer. Choice A is wrong because it invokes

\mu_k

after the block has stopped — but kinetic friction only applies to objects already in motion. If the block hasn't started sliding back yet,

\mu_k

is irrelevant to whether it will. Choice C makes a similar error, confusing the deceleration phase with the rest condition. The operative coefficient at the moment of stopping is

\mu_s

, not

\mu_k

. Choice D is wrong because

\mu_k \geq \tan\alpha

is never required for the block to remain stationary — that condition is simply redundant once

\mu_s \geq \tan\alpha

holds. Key takeaway: Always ask yourself which phase of motion a condition applies to. "Stays at rest" questions test static friction; "rate of deceleration" questions test kinetic friction. Don't mix them.

Question 3

A 15 kg block sits on a horizontal surface ( $\mu_s = 0.50$ , $\mu_k = 0.35$ ). A force $P = 60\text{ N}$ is applied at 20° above the horizontal. A separate horizontal force $Q = 20\text{ N}$ acts in the opposite horizontal direction. Use $g = 9.81\text{ m/s}^2$ .

Determine the friction force acting on the block and its state of motion.

Friction force $= 36.4\text{ N}$ (in the direction opposing $P$ 's horizontal component), and the block remains stationary, since the net applied horizontal force equals $P\cos 20° - Q = 56.4 - 20 = 36.4\text{ N}$ , which is less than $F_{s,max} = \mu_s N = 0.50(126.7) = 63.4\text{ N}$ . (correct answer)
Friction force $= 36.6\text{ N}$ (opposing net applied horizontal force), and the block slides because the net applied horizontal force of $56.4 - 20 = 36.4\text{ N}$ exceeds the maximum static friction of $\mu_s N$ , where $N = mg - P\sin 20° = 147.2 - 20.5 = 126.7\text{ N}$ , giving $F_{s,max} = 63.3\text{ N} > 36.4\text{ N}$ .
Friction force $= 44.3\text{ N}$ (opposing net horizontal force), and the block slides because applying $Q$ in the opposite direction increases the effective friction demand beyond $\mu_s N = 0.50(147.2) = 73.6\text{ N}$ ; the kinetic friction is $\mu_k N = 51.5\text{ N}$ , which cannot balance $P\cos 20°$ .
Friction force $= 63.4\text{ N}$ (kinetic, opposing motion), and the block slides in the direction of $P$ 's horizontal component, because the applied force $P\cos 20° = 56.4\text{ N}$ alone exceeds $\mu_k N = 0.35(126.7) = 44.3\text{ N}$ , and kinetic friction governs once slip is assumed.

Explanation: When analyzing friction problems, your first job is always to check whether the block actually moves — you do this by comparing the net applied horizontal force to the maximum static friction force. Never assume sliding; prove it. Start by finding the normal force. Since

P

is angled 20° above horizontal, its vertical component reduces the contact force:

N = mg - P\sin 20° = (15)(9.81) - 60\sin 20° = 147.2 - 20.5 = 126.7\text{ N}

. Now compute the maximum static friction:

F_{s,max} = \mu_s N = 0.50(126.7) = 63.4\text{ N}

. Next, find the net horizontal force the friction must resist:

P\cos 20° - Q = 56.4 - 20 = 36.4\text{ N}

. Since

36.4\text{ N} < 63.4\text{ N}

, static friction can hold the block, and the actual friction force equals exactly

36.4\text{ N}

— only as much as needed. Answer A is correct. Answer B correctly sets up the normal force and

F_{s,max}

, but then contradicts itself by claiming the block slides even though

36.4\text{ N} < 63.4\text{ N}

. The logic is inverted — a lower demand than capacity means no sliding. Answer C ignores the vertical component of

P

when computing

N

, using

mg

alone, which overstates the normal force and produces an incorrect

F_{s,max}

. It also fabricates a reason why

Q

increases friction demand. Answer D jumps straight to kinetic friction without first verifying that static friction is overcome — a classic procedural error. Kinetic friction only applies after you've confirmed sliding occurs. Strategy tip: Always follow the sequence — find

N

(accounting for angled forces), compute

F_{s,max}

, compare to net applied force, and only then decide if the block moves.

Question 4

A uniform block of mass $m$ , height $h$ , and width $b$ rests on a rough floor (static friction coefficient $\mu_s$ ). A horizontal force $P$ is applied at height $d$ from the floor. Assume $b < h$ .

For the block to slide rather than tip, which condition must hold?

$d < \dfrac{\mu_s h}{2}$ , because tipping requires the moment of $P$ about the tipping edge to exceed the restoring moment of gravity; equating the sliding and tipping critical forces introduces the block height $h$ through the overturning moment arm, yielding this height-based criterion.
$d > \dfrac{b}{2\mu_s}$ , because a higher application point generates a larger overturning moment, making tipping easier; sliding therefore only precedes tipping when $d$ is large enough that the tip-critical force drops below the slide-critical force.
$\mu_s > \dfrac{b}{2d}$ , because a higher friction coefficient makes sliding harder to initiate; when $\mu_s$ exceeds the geometric ratio $b/(2d)$ , the force required to slide the block exceeds the force required to tip it, so tipping occurs first.
$\mu_s < \dfrac{b}{2d}$ , because the block slides rather than tips when the friction limit $\mu_s mg$ is reached before the tipping moment condition $Pd = mg\,b/2$ is satisfied; setting $P_{\text{slide}} < P_{\text{tip}}$ gives $\mu_s mg < mg\,b/(2d)$ , i.e., $\mu_s < b/(2d)$ . (correct answer)

Explanation: When a block can either slide or tip, you need to compare two critical forces: the force that initiates sliding,

P_{\text{slide}} = \mu_s mg

, and the force that initiates tipping. Tipping occurs when

P

creates enough overturning moment about the bottom edge to overcome gravity's restoring moment. Balancing those moments gives

P_{\text{tip}} \cdot d = mg \cdot \frac{b}{2}

, so

P_{\text{tip}} = \frac{mgb}{2d}

. The block slides rather than tips when sliding is triggered first — that is, when

P_{\text{slide}} < P_{\text{tip}}

. Substituting:

\mu_s mg < \frac{mgb}{2d}

. Dividing both sides by

mg

gives

\mu_s < \frac{b}{2d}

, which is exactly answer D. Choice A introduces block height

h

into the criterion, but

h

never appears in the tipping moment equation — the relevant geometry is the half-width

b/2

and the application height

d

. This makes A physically incorrect regardless of its reasoning. Choice B states that sliding precedes tipping only when

d

is large, but a larger

d

actually lowers

P_{\text{tip}}

, making tipping easier, not harder. B has the inequality backwards and misidentifies which mode is favored. Choice C correctly identifies the relevant ratio

b/(2d)

but reverses the inequality. When

\mu_s > b/(2d)

, sliding requires more force than tipping, so the block tips first — the opposite of what the question asks. Study tip: On sliding-vs-tipping problems, always write out both critical forces explicitly, then set

P_{\text{slide}} < P_{\text{tip}}

and solve. Inequality direction mistakes (as in B and C) are the most common error on these problems.

Question 5

A block is pressed against a vertical wall by a horizontal force $F$ . The wall has static friction coefficient $\mu_s = 0.60$ and kinetic friction coefficient $\mu_k = 0.40$ with the block. The block has mass $m = 4\text{ kg}$ . Use $g = 9.81\text{ m/s}^2$ .

What is the minimum horizontal force $F$ required to prevent the block from sliding down the wall, and what is the direction of the friction force on the block at this minimum condition?

$F_{\min} = mg/(\mu_s) = 4(9.81)/0.60 \approx 65.4\text{ N}$ ; friction acts downward on the block, because the wall pushes back horizontally and friction must resist the tendency of $F$ to push the block upward along the wall surface.
$F_{\min} = mg/(\mu_k) = 4(9.81)/0.40 \approx 98.1\text{ N}$ ; friction acts upward on the block, because at the minimum force condition the block is on the verge of sliding and kinetic friction — not static — governs the threshold between sliding and not sliding.
$F_{\min} = mg/(\mu_s) = 4(9.81)/0.60 \approx 65.4\text{ N}$ ; friction acts upward on the block, because gravity pulls the block down, friction must act upward to maintain vertical equilibrium, and the normal force equals $F$ , so $F_{\min} = mg/\mu_s$ . (correct answer)
$F_{\min} = mg\mu_s = 4(9.81)(0.60) \approx 23.5\text{ N}$ ; friction acts upward on the block, because the maximum friction available is $\mu_s F$ and this must equal $mg$ ; solving $\mu_s F = mg$ directly gives $F = mg/\mu_s$ , but the formula is applied inverted since friction augments the pressing force.

Explanation: When a block is pressed against a vertical wall, you're dealing with two perpendicular force directions simultaneously. The horizontal force

F

creates the normal force on the wall, which in turn determines how much friction the wall can provide. Gravity acts downward, and friction must counteract it — so your first instinct should be: which direction does friction act, and what sets the normal force? Since gravity pulls the block down, the block's tendency is to slide downward, meaning static friction acts upward on the block. The normal force on the block from the wall equals

F

(the horizontal press). At the minimum force condition, static friction is maxed out:

f_s = \mu_s N = \mu_s F

. Setting this equal to the weight for vertical equilibrium:

\mu_s F = mg

, which gives

F_{\min} = mg/\mu_s = 4(9.81)/0.60 \approx 65.4 \text{ N}

. That's exactly what C states — making it correct. A is wrong on direction: gravity pulls the block down, so friction must act upward, not downward. There's no mechanism here pushing the block upward along the wall. B incorrectly applies kinetic friction. Kinetic friction describes surfaces already sliding relative to each other. The minimum-force threshold is a static equilibrium problem, so

\mu_s

governs, not

\mu_k

. Using

\mu_k

here is a classic conceptual error. D inverts the formula correctly in prose but mislabels it —

F = mg\mu_s

is the wrong equation (it would give an absurdly small force), despite the answer claiming to solve for

F_{\min} = mg/\mu_s

. Study tip: For wall-pressing problems, always identify the normal force source (here,

N = F

) and ask which direction the block tends to move — friction always opposes that tendency.

Question 6

Two blocks are connected by a horizontal cord passing over a frictionless pulley at the edge of a table. Block A (mass $m_A = 8\text{ kg}$ ) sits on the table; Block B (mass $m_B = 5\text{ kg}$ ) hangs vertically. The table surface has $\mu_s = 0.45$ and $\mu_k = 0.30$ . Use $g = 9.81\text{ m/s}^2$ .

A student claims: 'The system is on the verge of motion because the hanging weight (49.05 N) is nearly equal to the maximum static friction (35.3 N), and adding just 1 N to block B will definitely cause sliding.' Identify the critical flaw in this reasoning.

The student's comparison is fundamentally flawed: the hanging weight (49.05 N) is already greater than the maximum static friction (35.3 N), so the system is already sliding — it is not 'on the verge.' The student should have concluded the system is in kinetic motion, not impending static motion. (correct answer)
The student's numerical values are both computed incorrectly: the maximum static friction should use the total system weight, giving $\mu_s(m_A + m_B)g = 0.45(13)(9.81) = 57.3\text{ N}$ , which exceeds the hanging weight, so the system is actually safely in static equilibrium — not on the verge of motion.
The student confuses the cord tension with the weight of block B. The tension equals $m_B g$ only in static equilibrium or uniform motion; if the system is accelerating, the tension is less than $m_B g$ . Therefore, the correct driving force to compare against static friction is the cord tension, not the full weight of B.
The student neglects to verify that the cord remains taut: if static friction exceeds the hanging weight, the cord would go slack and the comparison between friction and hanging weight becomes meaningless, invalidating the entire analysis.

Explanation: When analyzing whether a system is in static equilibrium, on the verge of motion, or already moving, your first job is to compare the driving force against the maximum static friction force — and pay attention to which side is larger. Here, the driving force is the weight of block B:

m_B g = 5 \times 9.81 = 49.05 \text{ N}

. The maximum static friction resisting motion of block A is

\mu_s m_A g = 0.45 \times 8 \times 9.81 = 35.3 \text{ N}

. The student calculated both values correctly — but then misread what they mean. Since

49.05 \text{ N} > 35.3 \text{ N}

, the driving force already exceeds maximum static friction. The system is not "on the verge" of sliding — it is already sliding, and kinetic friction (

\mu_k = 0.30

) governs the motion. Answer A correctly identifies this: the student's fatal error was misinterpreting a clear case of kinetic motion as impending static motion. Answer B is wrong because maximum static friction depends only on the normal force acting on block A (which equals

m_A g

, not the total system weight). Block B hangs freely and contributes no normal force to the table surface. Answer C introduces a valid physics nuance about tension vs. weight during acceleration, but it does not address the actual flaw in the student's reasoning — the student never misidentified the driving force as tension. Answer D invents a concern about cord slack that is physically irrelevant here; the hanging block clearly keeps the cord taut. Study tip: Always ask "greater than, less than, or equal to?" after computing your friction and driving force values — the relationship between them tells you the entire story about the system's motion state.