The Workbook

1. \(N = mg = 50\) N. \(f_{\max} = \mu_sN = 20\) N; \(f_k = \mu_kN = 15\) N.

2. \(10\) N is less than \(20\) N, so the block stays at rest. Static friction equals the push: \(f = 10\) N, \(a = 0\).

3. \(25\) N exceeds \(20\) N, so the block slides and kinetic friction acts: \(f = 15\) N. \(a = \dfrac{25 - 15}{5} = 2\) m/s\(^2\).

1. The peak is \(20\) N: \(\mu_s = \dfrac{20}{50} = 0.4\).

2. The flat part is \(15\) N: \(\mu_k = \dfrac{15}{50} = 0.3\).

3. While the block is at rest the friction exactly balances the applied force, so \(f = F\). That is a line of slope \(1\), up to the limiting value.

Static friction takes whatever value is needed to keep the body at rest, from zero up to the limit \(\mu_sN\). It adjusts itself to the applied force. Once sliding starts the surfaces touch only briefly at their high points and grip less, so kinetic friction is smaller than the limiting static friction.

1. Steady speed means \(f_k = 6\) N. \(\mu_k = \dfrac{6}{20} = 0.3\).

2. With the pull removed only friction acts: \(a = \dfrac{6}{2} = 3\) m/s\(^2\) backward. \(s = \dfrac{u^2}{2a} = \dfrac{36}{6} = 6\) m.

The box is at rest, so friction balances the push: \(f = 30\) N. This must not exceed the limit: \(\mu_s \times 100 \ge 30\), so \(\mu_s \ge 0.3\). The exact value cannot be found.

1. Pull: \(N = 60 - F\sin 37^\circ = 60 - 0.6F\). Just moving: \(F\cos 37^\circ = \mu N\), so \(0.8F = 0.75(60 - 0.6F) = 45 - 0.45F\). \(1.25F = 45\), \(F = 36\) N.

2. Push: \(N = 60 + 0.6F\). \(0.8F = 0.75(60 + 0.6F) = 45 + 0.45F\), so \(0.35F = 45\) and \(F \approx 129\) N.

3. The pull's upward part reduces the normal force and so the friction. The push's downward part increases both. The same horizontal part must then beat a much larger friction.

A pull along a handle tilted upward has an upward component that lifts part of the roller's weight off the ground, reducing \(N\). A push along the handle presses the roller down and increases \(N\). Friction is proportional to \(N\), so pulling meets less friction.

1. Vertical: \(N = W - F\sin\theta\). Horizontal, at the point of moving: \(F\cos\theta = \mu N = \mu(W - F\sin\theta)\). So \(F(\cos\theta + \mu\sin\theta) = \mu W\) and \(F = \dfrac{\mu W}{\cos\theta + \mu\sin\theta}\).

2. \(\tan\theta = 0.75\) gives \(\theta = 37^\circ\). \(F = \dfrac{0.75 \times 60}{0.8 + 0.75 \times 0.6} = \dfrac{45}{1.25} = 36\) N. Question 6 used exactly this angle, so its \(36\) N is the least possible pull.

1. Along the slope the weight pulls with \(mg\sin 30^\circ = 10\) N, so friction must supply \(10\) N up the slope.

2. \(N = mg\cos 30^\circ = 10\sqrt{3}\) N. \(f_{\max} = 0.8 \times 10\sqrt{3} = 8\sqrt{3} \approx 13.9\) N.

3. \(10 < 13.9\), so the block stays. The friction is \(10\) N, only what is needed, not the full \(\mu_sN\).

At the angle of repose the friction is at its limit: \(mg\sin\theta_r = \mu_s mg\cos\theta_r\), so \(\tan\theta_r = \mu_s\).

\(\mu_s = 0.75\): \(\theta_r = 37^\circ\). \(\mu_s = \dfrac{1}{\sqrt{3}}\): \(\theta_r = 30^\circ\).

Down the slope: \(ma = mg\sin\theta - \mu_kmg\cos\theta\), so \(a = 10(0.6 - 0.25 \times 0.8) = 10 \times 0.4 = 4\) m/s\(^2\).

1. Going up, both gravity and friction act down the slope: \(a = 10(0.6 + 0.2) = 8\) m/s\(^2\). Distance \(= \dfrac{64}{16} = 4\) m.

2. Yes. At the top the block is at rest and \(\tan 37^\circ = 0.75\) exceeds \(\mu = 0.25\), so friction cannot hold it.

3. Coming down, \(a = 4\) m/s\(^2\) over \(4\) m: \(v = \sqrt{2 \times 4 \times 4} = 4\sqrt{2} \approx 5.7\) m/s, less than the \(8\) m/s it started with.

1. The block tends to slide down, so friction acts up the slope at its limit \(\mu N = 0.25 \times 16 = 4\) N. \(F + 4 = mg\sin 37^\circ = 12\), so \(F = 8\) N.

2. Now the block moves up, so friction acts down the slope: \(F = 12 + 4 = 16\) N.

3. Up the slope in (i), down the slope in (ii). Friction always opposes the sliding, or the tendency to slide.

\(a = g(\sin 30^\circ - \mu_k\cos 30^\circ) = 10\left(0.5 - \dfrac{1}{2\sqrt{3}} \times \dfrac{\sqrt{3}}{2}\right) = 10(0.5 - 0.25) = 2.5\) m/s\(^2\).

\(5 = \dfrac{1}{2} \times 2.5 \times t^2\), so \(t = 2\) s.

1. The friction peaks at \(37^\circ\), the angle of repose, with \(f = 12\) N \(= mg\sin 37^\circ\).

2. \(\mu_s = \tan 37^\circ = 0.75\).

3. \(N = mg\cos 37^\circ = 16\) N. \(\mu_k = \dfrac{8}{16} = 0.5\).

1. The greatest braking force is \(\mu mg\), so \(a = \mu g = 5\) m/s\(^2\).

2. \(s = \dfrac{u^2}{2a} = \dfrac{400}{10} = 40\) m.

3. No. The mass cancels: a heavier car has more friction and more inertia in the same proportion.

Only friction can accelerate the box: \(\mu_smg \ge ma\), so \(a_{\max} = \mu_sg = 4\) m/s\(^2\).

In the bus's frame the block feels a pseudo force \(5m\) forward. Friction can supply at most \(\mu mg = 4m\) backward, which is not enough. Net \(= m\), so the block slides forward relative to the bus at \(1\) m/s\(^2\).

\(\mu_s = \tan 30^\circ = \dfrac{1}{\sqrt{3}} \approx 0.58\).

1. The hanging weight \(40\) N exceeds the largest friction \(\mu mg = 0.5 \times 40 = 20\) N, so the system moves.

2. \(a = \dfrac{40 - 20}{4 + 4} = 2.5\) m/s\(^2\). \(T = 4(10 - 2.5) = 30\) N.

3. At rest the hanging weight must not exceed \(20\) N: \(m = 2\) kg.

Limit \(= 0.3 \times 40 = 12\) N. Push \(5\) N: friction \(5\) N. Push \(10\) N: friction \(10\) N. Push \(15\) N: the block slides and friction is the kinetic value, about \(12\) N (taking \(\mu_k \approx \mu_s\) here).

Vertically the block does not move: \(N + F\sin\theta = W\), so \(N = W - F\sin\theta\). As \(\theta\) grows, \(F\sin\theta\) grows and \(N\) falls, so the friction \(\mu N\) falls too. That is why a tilted pull helps, up to the best angle.

\(s = \dfrac{u^2}{2\mu g}\). Dry: \(\dfrac{400}{16} = 25\) m. Wet: \(\dfrac{400}{4} = 100\) m, four times as far.

Steady speed means no net force along the slope: \(mg\sin\theta = \mu_kmg\cos\theta\), so \(\mu_k = \tan\theta\). For \(37^\circ\): \(\mu_k = 0.75\).