Applications of Forces

Worked example 4.1 · Pushed horizontally

A particle of mass $5$ kg rests on a rough plane inclined at $30^\circ$ to the horizontal, with coefficient of friction $0.2$. It is held by a horizontal force $P$ in the vertical plane through a line of greatest slope. Find $P$ when the particle is on the point of moving (a) up the plane and (b) down the plane.

1. Perpendicular to the plane, $P$ presses the particle into it as well:

   $$R = 5g\cos 30^\circ + P\sin 30^\circ$$

   so the greatest friction is $0.2R = 8.487\ldots + 0.1P$. The weight’s component down the plane is $5g\sin 30^\circ = 24.5$ N.

2. (a) About to move up: friction acts down the plane. Along the plane,

   $$\begin{aligned} P\cos 30^\circ &= 24.5 + 0.2R \\ &= 32.98\ldots + 0.1P \end{aligned}$$

   so $0.766\ldots P = 32.98\ldots$ and $P = 43.1$ N (3 s.f.).

3. (b) About to slide down: friction acts up the plane. Along the plane,

   $$\begin{aligned} P\cos 30^\circ + 0.2R &= 24.5 \\ 0.966\ldots P + 8.487\ldots &= 24.5 \end{aligned}$$

   so $0.966\ldots P = 16.01\ldots$ and $P = 16.6$ N (3 s.f.).

Worked example 4.2 · The least coefficient of friction

A uniform rod $AB$ of length $2a$ and weight $W$ rests with $A$ on rough horizontal ground. It is held at $30^\circ$ to the ground by a force $P$ at $B$, acting at right angles to the rod in the vertical plane containing it. Find $P$ in terms of $W$, and the least possible value of the coefficient of friction.

1. Take moments about $A$, where both unknown ground forces act. $P$ is at right angles to the rod, $2a$ from $A$; the weight acts at the midpoint, a horizontal distance $a\cos 30^\circ$ from $A$:

   $$\begin{aligned} P \times 2a &= W \times a\cos 30^\circ \\ P &= \tfrac{\sqrt{3}}{4}W \end{aligned}$$

2. $P$ makes $30^\circ$ with the vertical. Resolve horizontally and vertically:

   $$\begin{aligned} F &= P\sin 30^\circ = \tfrac{\sqrt{3}}{8}W \\ R &= W - P\cos 30^\circ = \tfrac{5}{8}W \end{aligned}$$

3. The rod rests only if $F \le \mu R$, so $\mu \ge \dfrac{\sqrt{3}/8}{5/8} = \dfrac{\sqrt{3}}{5}$.

Worked example 4.3 · On the point of slipping

A uniform ladder $AB$, of length $4$ m and mass $12$ kg, rests with its foot $A$ on rough horizontal ground and its top $B$ against a smooth vertical wall. It makes $60^\circ$ with the ground and is on the point of slipping. Find the coefficient of friction between the ladder and the ground.

1. Vertically: $R = 12g$.

2. Moments about $A$. The wall’s force $S$ acts at $B$, a height $4\sin 60^\circ$ above $A$; the weight acts a horizontal distance $2\cos 60^\circ$ from $A$:

   $$\begin{aligned} S \times 4\sin 60^\circ &= 12g \times 2\cos 60^\circ \\ S &= 2\sqrt{3}\,g = 33.9 \text{ N} \end{aligned}$$

3. Horizontally $F = S$, and on the point of slipping $F = \mu R$:

   $$\mu = \frac{2\sqrt{3}\,g}{12g} = \frac{\sqrt{3}}{6} = 0.289 \text{ (3 s.f.)}$$

Worked example 4.4 · How far can he climb?

A uniform ladder of length $5$ m and mass $20$ kg rests on rough horizontal ground against a smooth vertical wall, at an angle $\alpha$ to the ground, where $\tan\alpha = \tfrac{4}{3}$. The coefficient of friction is $0.4$. A man of mass $80$ kg climbs the ladder. How far up the ladder can he climb before it slips?

1. Vertically: $R = 20g + 80g = 100g$, so friction can be at most $0.4 \times 100g = 40g$. On the point of slipping $F = 40g$, and horizontally the wall’s force is $S = F = 40g$.

2. $\sin\alpha = \tfrac{4}{5}$ and $\cos\alpha = \tfrac{3}{5}$. With the man $x$ m up the ladder, measured from the foot, the distances from the foot are:

   • the top: $5\sin\alpha = 4$ m up;
   • the ladder’s centre: $2.5\cos\alpha = 1.5$ m across;
   • the man: $x\cos\alpha = 0.6x$ m across.

3. Take moments about the foot and divide through by $g$:

   $$\begin{aligned} 40 \times 4 &= 20 \times 1.5 + 80 \times 0.6x \\ 160 &= 30 + 48x \\ x &= 2.71 \text{ m (3 s.f.)} \end{aligned}$$

Worked example 4.5 · Up a rough slope

Particle $A$, of mass $5$ kg, rests on a rough plane inclined at $\alpha$ to the horizontal, where $\tan\alpha = \tfrac{3}{4}$. The coefficient of friction is $\tfrac{1}{4}$. A light inextensible string, parallel to a line of greatest slope, joins $A$ over a smooth pulley at the top of the plane to particle $B$, of mass $6$ kg, which hangs freely. The system is released from rest. Find the acceleration and the tension in the string.

1. For $A$: $R = 5g\cos\alpha = 4g$, so friction is at most $\tfrac{1}{4} \times 4g = g$. The weight down the plane is $5g\sin\alpha = 3g$.

2. Does it move? $B$ pulls with $6g$, more than $3g + g = 4g$, so $B$ falls and $A$ moves up the plane, with limiting friction $g$ acting down it.

3. An equation of motion for each particle:

   $$\begin{aligned} 6g - T &= 6a \quad (B) \\ T - 3g - g &= 5a \quad (A) \end{aligned}$$

4. Adding eliminates $T$: $2g = 11a$, so $a = \tfrac{2g}{11} = 1.78\text{ m s}^{-2}$ (3 s.f.). Then $T = 6(g - a) = \tfrac{54g}{11} = 48.1$ N (3 s.f.).

Worked example 4.6 · When the string goes slack

In Example 4.5, $B$ starts $1.5$ m above the ground and does not rebound. Find how much further $A$ moves up the plane, and what happens to it next.

1. $B$ falls $1.5$ m from rest at $\tfrac{2g}{11}$: $v^2 = 2 \times \tfrac{2g}{11} \times 1.5 = \tfrac{6g}{11}$, so $v = 2.31\text{ m s}^{-1}$ (3 s.f.).

2. Then the string goes slack. With $T = 0$, the weight component $3g$ and the friction $g$ both act down the plane on $A$, so it decelerates at $\tfrac{4g}{5} = 7.84\text{ m s}^{-2}$.

3. From $v^2 = u^2 + 2as$ with final speed zero:

   $$\begin{aligned} 0 &= \tfrac{6g}{11} - 2 \times \tfrac{4g}{5}\,s \\ s &= \tfrac{15}{44} = 0.341 \text{ m (3 s.f.)} \end{aligned}$$

4. At rest, $\tan\alpha = 0.75$ is more than $\mu = 0.25$, so friction cannot hold $A$ on the plane.