Friction Applications - Statics and Dynamics

1. The rope can exert force either up or down the incline. Determine the range of rope tensions $T$ (in N) for which the crate remains stationary. Which of the following correctly identifies both the minimum tension required to prevent sliding down and the maximum tension before sliding up begins?
   Answer: $T_{\min} = W(\sin30° - \mu_s\cos30°) \approx 302 \text{ N}$ (rope pulls up the slope) and $T_{\max} = W(\sin30° + \mu_s\cos30°) \approx 1660 \text{ N}$ (rope pulls up the slope), since $\tan30° \approx 0.577 > \mu_s = 0.40$ means friction alone cannot prevent sliding. Whenever you see a friction-on-an-incline problem with a rope, your first instinct should be to check whether friction alone can hold the object. The test for this is simple: compare $\tan\theta$ to $\mu_s$. If $\tan\theta > \mu_s$, the incline is "too steep" for friction alone, and the rope must pull upward to prevent sliding down. Here, $\tan30° \approx 0.577 > \mu_s = 0.40$, so friction alone is insufficient — the crate will slide down without the rope. To find the minimum tension, balance forces along the incline with friction acting up the slope (opposing the tendency to slide down): $T_{\min} = W\sin30° - \mu_s W\cos30° = 200(9.81)(0.5) - 0.40(200)(9.81)(0.866) \approx 981 - 679 \approx 302 \text{ N}$. For the maximum tension, the crate now tends to slide up, so friction reverses direction and acts down: $T_{\max} = W\sin30° + \mu_s W\cos30° \approx 981 + 679 \approx 1660 \text{ N}$. This confirms A is correct. Choice B incorrectly claims $\mu_s > \tan30°$, which reverses the inequality — a critical arithmetic error that leads to a wrong physical conclusion. Choice C correctly finds $T_{\min}$ but botches $T_{\max}$ by using only the friction term, forgetting that the rope must also overcome gravity along the slope. Choice D ignores friction entirely when finding $T_{\min}$, as if the surface were frictionless. Study tip: Always draw a free-body diagram and explicitly ask "which way does friction act?" — it opposes impending motion, and that direction flips between the two limiting cases.
2. Which of the following correctly describes the condition for the wedge to be self-locking (i.e., it will not slide back out when $P$ is removed), and what is the minimum wedge angle $\phi$ at which self-locking is lost given $\mu_s = 0.25$?
   Answer: Self-locking requires the wedge angle to be less than the sum of friction angles at the two active sliding surfaces; with $\phi_f \approx 14.0°$ at each surface, the combined limit is $\approx 28°$. Since $15° < 28°$, the wedge is self-locking, but self-locking would be lost if the wedge angle exceeded $28°$. When analyzing wedge self-locking, the key insight is that two friction surfaces resist the wedge sliding back out simultaneously. As the wedge tries to back out, friction acts forward at both the wedge-floor interface and the wedge-block interface. This means the total resistance comes from both surfaces combined, not just one. The friction angle at each surface is $\phi_f = \arctan(\mu_s) = \arctan(0.25) \approx 14.0°$. For self-locking, the wedge angle $\alpha$ must satisfy $\alpha < \phi_{f1} + \phi_{f2}$, i.e., the wedge angle must be less than the sum of friction angles across the two active sliding surfaces. Here, that combined limit is $14.0° + 14.0° = 28°$. Since $15° < 28°$, the wedge is indeed self-locking, and self-locking would only be lost if $\alpha$ exceeded $28°$. This confirms C is correct. Answer A makes the error of comparing the wedge angle against a single friction angle (14°) rather than the sum, incorrectly concluding the wedge fails the self-locking condition. Answer B compounds this mistake by explicitly stating only the "most-critical" single surface matters — this ignores the additive nature of friction at multiple interfaces entirely. Answer D uses a simplified single-surface criterion $\mu_s > \tan\alpha$ without accounting for the two-surface geometry; it also incorrectly frames self-locking as independent of the number of active surfaces. A useful rule of thumb: count your friction surfaces. Every time a wedge backs out, it drags along multiple interfaces simultaneously, so always sum the friction angles from all active sliding contacts before comparing against the wedge angle.
3. When the horizontal force $F$ is directed into the slope (pushing the block against the incline) rather than parallel to it, which of the following correctly describes the effect on the normal force, the maximum static friction, and the equilibrium condition along the slope compared to the case of no horizontal force?
   Answer: The normal force increases to $N = mg\cos\theta + F\sin\theta$, raising maximum friction to $\mu_s(mg\cos\theta + F\sin\theta)$. The component of $F$ along the slope is $F\cos\theta$ directed down the slope, so no-slip requires $\mu_s(mg\cos\theta + F\sin\theta) \geq mg\sin\theta + F\cos\theta$. Whenever a force is applied to a block on an incline, your first move should be to resolve every force into slope-parallel and slope-perpendicular components, then apply equilibrium in each direction separately. Mixing up which component goes which direction is the most common error on these problems. A horizontal force $F$ directed into the slope has two components in slope coordinates: $F\sin\theta$ perpendicular to the surface (pressing the block harder against the incline) and $F\cos\theta$ parallel to the surface. To determine the direction of the parallel component, sketch the geometry: a horizontal vector pointing into a slope that rises to the right points partly downhill along the surface, so $F\cos\theta$ acts down the slope. The normal force therefore becomes $N = mg\cos\theta + F\sin\theta$, raising maximum static friction to $\mu_s(mg\cos\theta + F\sin\theta)$. For no-slip, this friction must resist both the gravitational down-slope component and the new $F\cos\theta$ driver, giving $\mu_s(mg\cos\theta + F\sin\theta) \geq mg\sin\theta + F\cos\theta$. That is exactly answer A, the correct choice. Answer B makes the right call on the normal force but incorrectly claims $F$ has no along-slope component — every force not perpendicular to the slope has a parallel component once you resolve it. Answer C wrongly assumes the normal force is unaffected by $F$; any force with a perpendicular-to-surface component absolutely changes $N$. Answer D reverses the sign of the perpendicular component (subtracting instead of adding) and flips the along-slope direction, both stemming from a faulty geometry sketch. Strategy tip: Always draw the incline, lay the force vector on it, and trace the perpendicular and parallel projections explicitly — sign errors almost always come from skipping this step.
4. An operator applies a 50 N force at the end of a 0.6 m lever arm to tighten the band. When the geometry is such that the tight side of the belt is connected directly to the lever pivot end (making the brake self-energizing), which of the following correctly computes the braking torque on the drum?
   Answer: The lever moment determines the slack-side tension $T_2 = 50 \times 0.6 / 0.3 = 100 \text{ N}$; because the tight side is connected to the pivot in the self-energizing arrangement, $T_1 = T_2 \cdot e^{\mu_k \beta} = 100 \times e^{0.30 \times 3\pi/2} \approx 411 \text{ N}$; braking torque $= (411 - 100) \times 0.3 \approx 93.3 \text{ N}\cdot\text{m}$. Whenever you see a band brake problem, your first job is identifying which end of the belt connects to the lever pivot — this determines whether the brake is self-energizing and which tension the lever controls directly. In a self-energizing configuration, the tight side connects to the lever pivot. This means friction amplifies the clamping force rather than working against it. Your lever moment equation gives you the slack-side tension $T_2$, and then friction multiplies it up to $T_1$. Taking moments about the lever pivot: $T_2 = \frac{50 \times 0.6}{0.3} = 100 \text{ N}$. The belt friction relationship for impending slip is $T_1 = T_2 \cdot e^{\mu_k \beta}$, where $\beta = \frac{3\pi}{2}$ rad. So $T_1 = 100 \times e^{0.30 \times 1.5\pi} \approx 100 \times 4.11 \approx 411 \text{ N}$. Braking torque is $(T_1 - T_2) \times r = 311 \times 0.3 \approx 93.3 \text{ N}\cdot\text{m}$. That's C. A is wrong because it ignores the belt friction relationship entirely — the exponential factor does not cancel; it's precisely what makes band brakes powerful. B correctly sets up the lever moment but misidentifies which end produces that tension: in a self-energizing brake, the lever controls $T_2$ (slack side), not $T_1$ (tight side). Assigning 100 N to $T_1$ and dividing gives the anti-energizing case. D uses the correct tension assignment but substitutes $\beta = \pi$ instead of the stated 270° = $\frac{3\pi}{2}$ — a unit/conversion error. A reliable strategy: always sketch the drum rotation direction, mark which belt end is "peeling off" (slack) versus being "dragged in" (tight), and trace from there to the lever to identify which tension the applied force directly sets.
5. As $P$ is gradually increased from zero, which surface slips first and at what value of $P$ does slipping begin? (Assume the upper block is not separately restrained.)
   Answer: The block-block surface slips first at $P = 175 \text{ N}$. The inter-block friction limit is $\mu_s W = 0.35 \times 500 = 175 \text{ N}$, while the floor can sustain up to $\mu_s(2W) = 350 \text{ N}$. Because $P$ is applied only to the lower block, the upper block can receive horizontal force only through the block-block interface; that interface reaches its friction limit before the floor does. When a force is applied to only one body in a stacked system, your first job is to identify which interface must transmit force to the other body — because that interface will be the friction "bottleneck." Here, the upper block has no applied force. The only way it can experience any horizontal force is through friction at the block-block interface. So as you increase $P$, the upper block-lower block interface must transmit whatever horizontal force keeps the upper block accelerating (or attempting to move) with the lower block. Meanwhile, the floor must resist the entire applied force $P$. Comparing the two friction limits tells you which surface gives way first. The block-block interface supports a normal force equal to the upper block's weight: $N_{BB} = W = 500 \text{ N}$, giving a friction limit of $\mu_s W = 0.35 \times 500 = 175 \text{ N}$. The floor supports both blocks: $N_{floor} = 2W = 1000 \text{ N}$, with a limit of $\mu_s(2W) = 350 \text{ N}$. The block-block interface reaches its limit at $P = 175 \text{ N}$ — well before the floor does. Answer B is correct. Answer A is wrong because it assumes no horizontal force is transmitted through the block-block interface, which violates Newton's second law for the upper block — something must act on it horizontally. Answer C is wrong because identical $\mu_s$ values don't mean identical friction limits; the limits depend on normal force, which differs at each interface. Answer D correctly identifies the block-block surface as the first to slip but miscalculates the threshold, confusing the floor's friction limit with the block-block limit. The key study tip: always draw a free-body diagram for each body separately. The interface that has the smaller normal force — and thus the smaller friction limit — will slip first, regardless of where the external force is applied.
6. The load is on the verge of slipping downward (tight side slipping relative to drum). An engineer proposes adding a second identical drum in series — the belt wraps $180°$ around the first drum, then $180°$ around a second drum, then to the operator's hand. Both drums are fixed (non-rotating) and have the same $\mu_s = 0.40$. What minimum operator force $T_{\text{hand}}$ is now needed to hold the 200 N load on the verge of slipping?
   Answer: $T_{\text{hand}} = 200 / e^{0.40 \times 2\pi} \approx 200 / 12.35 \approx 16.2 \text{ N}$, because the two drums act in series and the total effective angle of wrap is $2 \times 180° = 360°$ ($2\pi$ rad), multiplying the friction effect. When a belt wraps around multiple fixed drums in series, think of each drum as an independent friction brake — the belt tension leaving one drum becomes the input tension entering the next. This is the key insight: the exponential friction factors multiply, not add. Recall the capstan (belt-friction) equation for a single drum: $T_{\text{tight}} / T_{\text{slack}} = e^{\mu_s \beta}$. For your first drum with $\mu_s = 0.40$ and $\beta = \pi$, the ratio is $e^{0.40\pi} \approx 3.514$. So the tension drops from 200 N to $200 / e^{0.40\pi}$ after drum one. That reduced tension then enters drum two as the new "tight side," and the same equation applies again — dividing by another $e^{0.40\pi}$. The total reduction is $e^{0.40\pi} \times e^{0.40\pi} = e^{0.40 \times 2\pi}$, giving $T_{\text{hand}} = 200 / e^{0.8\pi} \approx 200 / 12.35 \approx 16.2 \text{ N}$. This is answer B, and it's equivalent to saying the total wrap angle is $2\pi$ — a useful shortcut when drums have identical $\mu_s$. A is wrong because the second drum is not redundant — each drum independently reduces tension, so adding drums always helps. C incorrectly divides the exponential factor by 2 rather than squaring it, confusing "sharing" a force with chaining a ratio. D makes the same arithmetic error as C but frames it differently — friction reductions compound multiplicatively through $e^{\mu\beta}$, never additively. Study tip: When you see drums or capstans in series, always multiply their exponential factors (equivalently, add the wrap angles). The capstan equation is a ratio, so series arrangements behave like compounding interest — exponential, not linear.
7. The ship's engine exerts a steady pull on the tight end of the rope. What is the maximum load the rope can hold, and if the worker doubles the number of turns to $n = 6$ while keeping their holding force at 100 N, by what factor does the maximum holdable load increase?
   Answer: With $n = 3$ turns, $\beta = 6\pi$ rad; $T_1 = 100 \cdot e^{0.25 \times 6\pi} \approx 11{,}100 \text{ N}$. Adding 3 more turns increases the wrap angle by $6\pi$ rad, so the additional multiplicative factor is $e^{0.25 \times 6\pi} \approx 111$, and the maximum holdable load increases by a factor of approximately 111. Whenever you see a capstan or rope-wrap problem, your anchor concept is the capstan (Euler-Eytelwein) equation: $T_\text{high} = T_\text{low} \cdot e^{\mu_s \beta}$, where $\beta$ is the total wrap angle in radians. The critical conversion to internalize: **one full turn = $2\pi$ radians**, not $\pi$. With $n = 3$ turns, $\beta = 3 \times 2\pi = 6\pi$ rad. Plugging in: $T_1 = 100 \cdot e^{0.25 \times 6\pi} \approx 100 \times 111 \approx 11{,}100 \text{ N}$. When the worker doubles to $n = 6$ turns, $\beta' = 12\pi$ rad. The new load is $T_1' = 100 \cdot e^{0.25 \times 12\pi}$. Because exponent rules give $e^{0.25 \times 12\pi} = e^{0.25 \times 6\pi} \cdot e^{0.25 \times 6\pi}$, the load multiplies by an additional factor of $e^{0.25 \times 6\pi} \approx 111$. Answer D captures this correctly. A fails immediately by using $2\pi \times n / 2 = n\pi$—treating one turn as $\pi$ rad instead of $2\pi$ rad. This halves every exponent and dramatically underestimates the friction effect. B correctly computes the initial load but then claims doubling the turns only doubles the holdable force. This confuses linear scaling with exponential scaling—the relationship is multiplicative through the exponent, not additive. C gets both loads right numerically ($\approx 11{,}100$ N and $\approx 1{,}230{,}000$ N) but then describes the increase as additive rather than multiplicative. The question asks for a factor, and $T_1'/T_1 \approx 111$ is the correct frame. Your study tip: always convert turns to radians using $2\pi$ per turn, and remember that adding wraps multiplies the holdable load exponentially—doubling turns doesn't double the load, it squares the load ratio.
8. After the machine is leveled and the driving force is removed, an engineer claims the wedge will remain in place without any locking device. A second engineer disputes this, arguing that a small tap will cause the wedge to slide out. Which analysis is correct, and what is the key criterion that resolves the dispute?
   Answer: The first engineer is correct. The self-locking criterion for a two-surface wedge is $\alpha < \phi_{f1} + \phi_{f2}$; with $\phi_{f1} = \phi_{f2} \approx 8.53°$, the combined limit is $\approx 17.06°$, which exceeds the 5° wedge angle, so the wedge is self-locking and will remain in place without external force. Wedge self-locking questions hinge on one critical insight: when a wedge has two friction-active surfaces, the friction resistance from both interfaces combines to resist back-sliding. The correct self-locking criterion is therefore $\alpha < \phi_{f1} + \phi_{f2}$, not simply $\alpha < \phi_f$ at a single surface. Here, both surfaces share the same friction angle $\phi_{f1} = \phi_{f2} = \arctan(0.15) \approx 8.53°$. The combined limit is $8.53° + 8.53° = 17.06°$. Since the wedge angle $\alpha = 5°$ is well below this threshold, the wedge is self-locking — the first engineer is correct, and D is the right answer. Choice A reaches the correct conclusion but for the wrong reason. Checking only the "single most-loaded interface" ignores that back-sliding must overcome friction at both surfaces simultaneously. The reasoning is incomplete even if the conclusion happens to be right here. Choice B inverts the logic entirely. It correctly notes that $\phi_f > \alpha$ at each individual surface, but then incorrectly claims the two surfaces combine against self-locking. In reality, friction forces from multiple surfaces combine in favor of locking. Choice C introduces a fictitious "effective friction coefficient" that gets halved when the driving force is removed. No such halving occurs — friction capacity at each surface remains $\mu_s$ regardless of load direction reversal. This is a fabricated rule. Study tip: Whenever you see a multi-surface wedge, always sum the friction angles from all sliding interfaces: self-locking requires $\alpha < \sum \phi_{fi}$. Applying the single-surface rule to a two-surface problem is the most common trap on wedge questions.
9. An engineer proposes increasing the power transmitted by replacing the flat belt with a V-belt whose groove half-angle is $\alpha = 18°$, keeping all other parameters ($\mu_s$, $\beta$, $T_1$) identical. Which of the following correctly computes the new slack-side tension $T_2$ and describes the resulting change in net tangential force on the pulley?
   Answer: The effective friction coefficient becomes $\mu_{\text{eff}} = \mu_s/\sin\alpha = 0.35/\sin18° \approx 1.132$, so $T_1/T_2 = e^{1.132 \times 2.79} \approx e^{3.16} \approx 23.6$. The slack-side tension drops to $T_2 \approx 34 \text{ N}$, raising the net tangential force to $\approx 766 \text{ N}$ — a large increase over the flat-belt value of 499 N. Whenever you see a question comparing flat belts to V-belts, the key insight is understanding why V-belts transmit more power: the groove geometry creates a wedging action that amplifies the normal force on the belt, effectively increasing friction without changing $\mu_s$. For a V-belt, the belt wedges into a groove with half-angle $\alpha$, so the two groove faces each push on the belt with a normal force $N$. The vertical components must support the radial belt load, giving a resultant normal force proportional to $1/\sin\alpha$ rather than simply the radial load itself. This amplifies friction, so the modified capstan equation uses an effective friction coefficient $\mu_{\text{eff}} = \mu_s/\sin\alpha$. With $\alpha = 18°$: $\mu_{\text{eff}} = 0.35/\sin 18° \approx 0.35/0.309 \approx 1.132$. The tension ratio becomes $T_1/T_2 = e^{1.132 \times 2.79} \approx e^{3.16} \approx 23.6$, so $T_2 = 800/23.6 \approx 34 \text{ N}$, and the net tangential force is $800 - 34 \approx 766 \text{ N}$. This is choice A, the correct answer. Choice B incorrectly multiplies $\mu_s$ by $\sin\alpha$ instead of dividing, which would actually reduce friction — the opposite of what a V-groove does. Choice C ignores the groove geometry entirely, wrongly treating the V-belt as a flat belt. Choice D divides by $\cos\alpha$ instead of $\sin\alpha$, confusing the trigonometric relationship; $\cos 18°$ is close to 1, so it yields almost no improvement over the flat belt. Study tip: Memorize the V-belt formula as $\mu_{\text{eff}} = \mu_s/\sin\alpha$ — the small half-angle (small $\sin\alpha$) means large effective friction. The shallower the groove, the greater the wedging amplification.

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