Forces, Density and Pressure: Question 3

Worked solution

Setting up the problem

Place the foot of the ladder at A, with the wall vertical and the ground horizontal. Since the ladder makes $60^\circ$ with the ground, a point a distance $s$ along the ladder from A is a horizontal distance $s\cos60^\circ$ from A and a height $s\sin60^\circ$ above A.

The forces on the ladder are: its own weight $200\text{ N}$, acting at its midpoint, $2.0\text{ m}$ from A along the ladder; the person’s weight $600\text{ N}$, acting $3.0\text{ m}$ from A along the ladder; the wall’s normal reaction $N$ on the ladder at B, which is horizontal (a smooth wall exerts no friction); and, at A, the ground’s normal reaction $N_g$ (vertical) and frictional force $F$ (horizontal, acting towards the wall to stop the foot sliding away from it).

Part (a): Normal reaction from the wall

Take moments about A. The ground’s normal reaction $N_g$ and friction $F$ both act at A, so neither has any perpendicular distance from A and neither appears in the moments equation. This is exactly why A is the useful pivot for finding $N$.

The perpendicular distance from A to the horizontal line of action of $N$ is the height of B above A: $4.0\sin60^\circ = 4.0\times\frac{\sqrt{3}}{2} = 2.0\sqrt{3}\text{ m}$

The perpendicular distance from A to the vertical line of action of each weight is its horizontal distance from A: $\text{ladder's weight: } 2.0\cos60^\circ = 2.0\times0.5 = 1.0\text{ m}$ $\text{person's weight: } 3.0\cos60^\circ = 3.0\times0.5 = 1.5\text{ m}$

The weight of the ladder and the weight of the person both tend to rotate the ladder about A so that B slides down the wall; the wall’s reaction $N$, acting horizontally at B, opposes this. By the principle of moments: $N\times2.0\sqrt{3} = (200\times1.0)+(600\times1.5)$ $N\times2.0\sqrt{3} = 200+900 = 1100$ $N = \frac{1100}{2.0\sqrt{3}} = \frac{550}{\sqrt{3}} = \frac{550\sqrt{3}}{3}$ $N \approx 317.5\text{ N} \approx 318\text{ N (3 s.f.)}$

Part (b): Frictional and normal reaction forces from the ground

Horizontally ($\sum F_x = 0$): the only other horizontal force on the ladder is the wall’s reaction $N$, which pushes the ladder away from the wall. The ground’s friction $F$ must balance this, acting towards the wall: $F = N = \frac{550\sqrt{3}}{3} \approx 317.5\text{ N} \approx 318\text{ N (3 s.f.), directed horizontally towards the wall}$

Vertically ($\sum F_y = 0$): the wall is smooth, so it exerts no vertical force at all. The ground’s normal reaction $N_g$ must alone support the total weight: $N_g = 200+600 = 800\text{ N, upward}$

Part (c): Resultant force from the ground

Combine the two perpendicular components $F$ and $N_g$ with Pythagoras’ theorem, keeping $F$ in its exact unrounded form so no rounding error is carried through: $F^2 = \left(\frac{550\sqrt{3}}{3}\right)^2 = \frac{550^2\times3}{9} = \frac{302500}{3} \approx 100{,}833\text{ N}^2$ $R = \sqrt{F^2+N_g^2} = \sqrt{100{,}833+640{,}000} = \sqrt{740{,}833}$ $R \approx 860.7\text{ N} \approx 861\text{ N (3 s.f.)}$

The direction, measured above the horizontal: $\theta = \arctan\left(\frac{N_g}{F}\right) = \arctan\left(\frac{800}{550\sqrt{3}/3}\right) = \arctan\left(\frac{16\sqrt{3}}{11}\right) \approx \arctan(2.519) \approx 68.4^\circ\text{ (3 s.f.)}$

So the ground exerts a force of about $861\text{ N}$ on the ladder, at about $68.4^\circ$ above the horizontal, directed towards the wall.

Final answers

• (a) $N \approx \boxed{318}\text{ N}$
• (b) Frictional force $F \approx \boxed{318}\text{ N}$ (towards the wall); normal reaction $N_g = \boxed{800}\text{ N}$ (upward)
• (c) Resultant force $\approx \boxed{861}\text{ N}$, at about $\boxed{68.4}^\circ$ above the horizontal, directed towards the wall