Statics and Dynamics Quiz: Friction Applications

What this quiz covers

This quiz focuses on Friction Applications, 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.

Question 1

1. $T_{\text{hand}} = 200 / e^{0.40 \times \pi} \approx 200 / 3.514 \approx 56.9 \text{ N}$, the same as with one drum, because the second drum is redundant once the belt is already on the verge of slipping.
2. $T_{\text{hand}} = 200 / e^{0.40 \times 2\pi} \approx 200 / 12.35 \approx 16.2 \text{ N}$, 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. (correct answer)
3. $T_{\text{hand}} = 200 / (2 \times e^{0.40 \times \pi}) \approx 200 / 7.03 \approx 28.5 \text{ N}$, because two drums share the holding friction equally, so each contributes half the exponential factor and the combined effect is twice the single-drum friction force.
4. $T_{\text{hand}} = 200 / (e^{0.40\pi} + e^{0.40\pi}) = 200 / (2 \times 3.514) \approx 28.5 \text{ N}$, because the belt tension drops by $e^{\mu\beta}$ at the first drum and then again by the same factor at the second drum, and these drops add rather than multiply.

Question 2

1. Self-locking requires the wedge angle to be less than the friction angle at each interface; with $\mu_s = 0.25$ the friction angle is $\phi_f = \arctan(0.25) \approx 14.0°$, so a 15° wedge just fails the self-locking condition and will slide back under zero applied load.
2. Self-locking requires the wedge angle to be less than the friction angle at the single most-critical contact surface; with $\phi_f \approx 14.0°$, the 15° wedge exceeds this limit and is therefore not self-locking, regardless of how many surfaces are active.
3. 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 $28°$. (correct answer)
4. Self-locking is independent of the number of active friction surfaces; the criterion is simply $\mu_s > \tan\alpha$. Since $\tan15° \approx 0.268 > 0.25$, the condition is narrowly violated and the wedge will slide back when $P$ is removed.

Question 3

1. The floor surface slips first at $P = 350 \text{ N}$. Since $P$ is applied to the lower block and the upper block is simply resting on it, the block-block interface carries no horizontal force; the full applied force $P$ acts against floor friction alone, which reaches its limit $\mu_s(2W) = 350 \text{ N}$ when the system slides as a unit.
2. 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. (correct answer)
3. Both surfaces slip simultaneously at $P = 350 \text{ N}$. Because $\mu_s$ is identical at both interfaces and the blocks are identical, the ratio of friction force to normal force is the same at each surface, so both surfaces reach their limit at the same applied force.
4. The block-block surface slips first, but at $P = 350 \text{ N}$, not 175 N. The upper block contributes its weight to the normal force at the block-block interface ($N = W = 500 \text{ N}$), but the friction limit there must equal the floor friction limit ($350 \text{ N}$) for the system to remain consistent, so slipping begins only when $P$ reaches 350 N.

Question 4

1. $T_{\min} = W(\sin30° - \mu_s\cos30°) \approx 302 \text{ N}$ (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. (correct answer)
2. $T_{\min} = 0 \text{ N}$ because $\mu_s = 0.40 > \tan30° \approx 0.577$, meaning friction alone holds the crate without any rope assistance, and $T_{\max} = W(\sin30° + \mu_s\cos30°) \approx 1660 \text{ N}$ before sliding up begins.
3. $T_{\min} = W(\sin30° - \mu_s\cos30°) \approx 302 \text{ N}$ (rope pulls up the slope) and $T_{\max} = W\mu_s\cos30° \approx 681 \text{ N}$, because once the rope tension exceeds the maximum static friction component, the crate begins to slide upward.
4. Since the crate tends to slide down, the rope must pull up the slope; the required tension is simply the net down-slope force with no friction contribution: $T_{\min} = W\sin30° = 981 \text{ N}$, and the maximum before sliding up is $T_{\max} = W(\sin30° + \mu_s\cos30°) \approx 1660 \text{ N}$.

Question 5

1. 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 $\mu_s(mg\cos\theta + F\sin\theta) \geq mg\sin\theta + F\cos\theta$. (correct answer)
2. The normal force increases to $N = mg\cos\theta + F\sin\theta$, raising maximum friction to $\mu_s(mg\cos\theta + F\sin\theta)$. However, $F$ is horizontal and the slope reaction is perpendicular to the surface, so $F$ has no component along the slope; no-slip requires only $\mu_s(mg\cos\theta + F\sin\theta) \geq mg\sin\theta$.
3. The horizontal force does not affect the normal force because contact pressure is set by the block's weight alone; friction remains $\mu_s mg\cos\theta$. However, $F\cos\theta$ acts as an additional down-slope driving force, making equilibrium harder to maintain.
4. Resolving $F$ into slope coordinates, the along-slope component $F\cos\theta$ acts up the slope (opposing the tendency to slide down) while the perpendicular component $F\sin\theta$ reduces the normal force. Therefore $N = mg\cos\theta - F\sin\theta$ and the net down-slope force decreases to $mg\sin\theta - F\cos\theta$.

Question 6

1. With $n = 3$ turns, $\beta = 3\pi$ rad (one turn = $\pi$ rad); $T_1 = 100 \cdot e^{0.25 \times 3\pi} \approx 100 \times 10.6 = 1{,}060 \text{ N}$. Doubling to $n = 6$ gives $\beta = 6\pi$ rad, so the load increases by a factor of $e^{0.25 \times 3\pi} \approx 10.6$.
2. With $n = 3$ turns, $\beta = 6\pi$ rad; $T_1 = 100 \cdot e^{0.25 \times 6\pi} \approx 11{,}100 \text{ N}$. Doubling the turns doubles the angle, so $T_1' = 2T_1 \approx 22{,}200 \text{ N}$; the load increases by a factor of exactly 2, since the exponential argument scales linearly with turns.
3. With $n = 3$ turns, $\beta = 6\pi$ rad; $T_1 = 100 \cdot e^{0.25 \times 6\pi} \approx 11{,}100 \text{ N}$. Doubling to $n = 6$ gives $\beta' = 12\pi$ rad, so $T_1' = 100 \cdot e^{0.25 \times 12\pi} \approx 1{,}230{,}000 \text{ N}$; the net increase is additive: $T_1' - T_1 \approx 1{,}219{,}000 \text{ N}$, not a simple multiplicative factor.
4. 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. (correct answer)

Question 7

1. 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. (correct answer)
2. The effective friction coefficient becomes $\mu_{\text{eff}} = \mu_s \cdot \sin\alpha \approx 0.35 \times 0.309 \approx 0.108$, so $T_1/T_2 = e^{0.108 \times 2.79} \approx e^{0.302} \approx 1.35$. The slack-side tension rises to $T_2 \approx 593 \text{ N}$, substantially decreasing the net tangential force to only $\approx 207 \text{ N}$.
3. The V-belt groove wedges the belt radially but does not alter the friction relationship in the capstan equation; the effective friction coefficient remains $\mu_s = 0.35$, so $T_2 \approx 301 \text{ N}$ and the net tangential force is unchanged at $\approx 499 \text{ N}$.
4. The effective friction coefficient becomes $\mu_{\text{eff}} = \mu_s/\cos\alpha = 0.35/\cos18° \approx 0.368$, so $T_1/T_2 = e^{0.368 \times 2.79} \approx e^{1.027} \approx 2.79$. The slack-side tension drops to $T_2 \approx 287 \text{ N}$, giving a marginally increased net tangential force of $\approx 513 \text{ N}$.

Question 8

1. The 50 N applied force acts directly as the net belt force on the drum; braking torque $= F \times r = 50 \times 0.3 = 15 \text{ N·m}$, since in a self-energizing configuration the lever arm and exponential friction factor cancel and the drum radius alone sets the torque.
2. Taking moments about the lever pivot gives tight-side tension $T_1 = 50 \times 0.6 / 0.3 = 100 \text{ N}$; this is the larger tension, so $T_2 = T_1 / e^{\mu_k \beta} = 100/e^{0.30 \times 3\pi/2} \approx 100/4.11 \approx 24.3 \text{ N}$; braking torque $= (T_1 - T_2) \times r \approx 75.7 \times 0.3 \approx 22.7 \text{ N·m}$.
3. 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 $= (411 - 100) \times 0.3 \approx 93.3 \text{ N·m}$. (correct answer)
4. The lever moment gives $T_2 = 50 \times 0.6 / 0.3 = 100 \text{ N}$; the wrap angle for 270° is $\beta = \pi$ rad (half-turn equivalent for a self-energizing brake), so $T_1 = 100 \times e^{0.30 \times \pi} \approx 100 \times 2.57 \approx 257 \text{ N}$; braking torque $= (257 - 100) \times 0.3 \approx 47.1 \text{ N·m}$.

Question 9

1. The first engineer is correct. Since the wedge angle $\alpha = 5°$ is less than the friction angle $\phi_f \approx 8.53°$ at the single most-loaded interface, self-locking is guaranteed at every individual contact, and the wedge cannot slide back under any load less than the original driving force.
2. The second engineer is correct. Because the wedge has two friction-active surfaces, the self-locking condition requires $\alpha < \phi_f$ at each individual surface; since $\phi_f \approx 8.53° > 5°$, self-locking holds at each surface individually, but the two surfaces together produce a net resultant that overcomes friction, so the wedge slides out spontaneously.
3. The second engineer is correct. The net friction force on the wedge upon removal of $P$ reverses direction, and since the reversed friction force must now act against the load rather than with it, the effective friction coefficient is halved, leaving $\mu_{eff} = 0.075$, which is insufficient to prevent back-sliding for any wedge angle greater than $\arctan(0.075) \approx 4.3°$.
4. 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. (correct answer)