Dry Friction Model - Statics and Dynamics

1. Determine the friction force acting on the block and its state of motion.
   Answer: 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 $F_{s,max} = \mu_s N = 0.50(126.7) = 63.4\text{ N}$. 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.
2. 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 $P$?
   Answer: $P$ 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. 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.
3. 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?
   Answer: $F_{\min} = mg/(\mu_s) = 4(9.81)/0.60 \approx 65.4\text{ N}$; 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 $F_{\min} = mg/\mu_s$. 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.
4. 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.
   Answer: 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. 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.
5. 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)?
   Answer: $\mu_s \geq \tan\alpha$, 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. 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.
6. For the block to slide rather than tip, which condition must hold?
   Answer: $\mu_s < \dfrac{b}{2d}$, 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., $\mu_s < b/(2d)$. 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.

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