Friction Questions and Answers

Answer: A

The block starts to slide when F = \(\mu W\).

Spring \(k_{2}\) transmits the full pull F to the block, whatever its stiffness, since it is massless and in series with the applied force. Spring \(k_{1}\) is tied to the wall and is undeformed at the start; it can only exert a force once the block moves and stretches it.

So while the block is at rest, static friction alone resists F, and slipping begins when F reaches the limiting friction \(\mu W\). The options with stiffness ratios would apply only if the block had already been displaced, so that \(k_{1}\) were carrying part of the load.

Answer: B

The minimum effort is \(W\left(\frac{\mu_{s} + \tan\theta}{1 - \mu_{s}\tan\theta}\right)\).

When upward motion impends, friction \(\mu_{s}N\) acts down the slope. Resolve along and normal to the incline for the horizontal force P:

\[ \begin{aligned} P\cos\theta &= W\sin\theta + \mu_{s}N \\ N &= W\cos\theta + P\sin\theta \end{aligned} \]

Substituting and dividing by \(\cos\theta\) gives \(P(1 - \mu_{s}\tan\theta) = W(\tan\theta + \mu_{s})\), the keyed expression (equal to \(W\tan(\theta + \phi)\) with \(\tan\phi = \mu_{s}\)).

The expressions with \(\mu_{s} - \tan\theta\) belong to a block about to slide down, and the one with \(1 + \mu_{s}\tan\theta\) in the denominator gets the sign of the normal-force contribution of P wrong.

Answer: A

Equilibrium is impossible for a spring force of 30 N.

The incline rises 3 m over 4 m, so \(\sin\theta = 0.6\) and \(\cos\theta = 0.8\). The spring pulls the mass up the slope. Along the incline:

\[ \begin{aligned} mg\sin\theta &= 100 \times 0.6 = 60\ \text{N (down the slope)} \\ \mu N &= 0.25 \times 100 \times 0.8 = 20\ \text{N} \end{aligned} \]

Friction can act either way, up to 20 N, so the spring force S must satisfy \(60 - 20 \le S \le 60 + 20\), that is \(40 \le S \le 80\) N. 45, 60 and 75 N all lie in this range; 30 N is too small to hold the mass even with full friction helping.

Answer: B

The correct free body diagram is (B).

A free body diagram shows the block alone, with every external force: W at the centre of gravity, P through the centre of gravity, friction F at the base opposing P, and the normal reaction N. Horizontal balance gives F = P and vertical balance N = W.

F acts at the base, below the line of P, so P and F form a couple that tends to tip the block forward. To balance it, N must act at a point shifted forward of W's line, in the direction of P, by \(x = Ph/W\) for P at height h. Diagram (B) shows this shift.

In (A) N and W are collinear, leaving the P-F couple unbalanced. In (C) N is shifted the wrong way. (D) draws the supporting surface, which a free body diagram must leave out, and has N in line with W.

Answer: A

The minimum force is about 862.2 N.

As W pushes Q toward the wall, the inclined face of Q lifts P up the frictionless wall. Relative to Q, P slides up the incline, so friction on P (0.2N) acts down the incline. Take N as the normal force between the blocks and \(\theta = 30^{\circ}\).

Vertical equilibrium of P, weight 100 × 9.81 = 981 N:

\[ \begin{aligned} N\cos\theta - 0.2N\sin\theta &= 981 \\ N(0.866 - 0.1) &= 981 \\ N &= 1280.7\ \text{N} \end{aligned} \]

Horizontally, Q has only W on one side and the horizontal parts of N and the friction from P on the other (the floor is frictionless):

\[ \begin{aligned} W &= N\sin\theta + 0.2N\cos\theta \\ &= 1280.7(0.5 + 0.1732) = 862.2\ \text{N} \end{aligned} \]

Q's own weight acts vertically and is carried by the floor, so it does not enter W. Leaving out friction gives \(981\tan 30^{\circ} = 566\) N, which is not offered; the larger options do not satisfy the two equilibrium equations.