Forces and Equilibrium: Question 10

Worked solution

Setting up the forces

The block has weight $W = mg = 5 \times 10 = 50\text{ N}$, acting vertically downwards. It is convenient to resolve perpendicular to the plane and along the plane (parallel to the line of greatest slope). Both the weight and the horizontal force $H$ have components in each of these directions, since neither acts purely along one of the chosen axes.

Part (a): Showing $R = mg\cos32^\circ + H\sin32^\circ$

Resolving the weight relative to the incline gives a component $mg\cos32^\circ$ perpendicular to the plane (pressing the block into the surface) and $mg\sin32^\circ$ along the plane (down the slope). This is the standard resolution used for any object on this incline.

The force $H$ is horizontal, not vertical, so relative to the incline its components are the other way round: a component $H\cos32^\circ$ along the plane (up the slope, since $H$ is directed to push the block towards and up the incline) and a component $H\sin32^\circ$ perpendicular to the plane, also pressing the block into the surface (this follows from resolving a horizontal force against axes tilted at $32^\circ$ to the horizontal).

Since the block does not move off the plane, resolving perpendicular to the plane (taking “away from the surface” as positive): $R - mg\cos32^\circ - H\sin32^\circ = 0$ $R = mg\cos32^\circ + H\sin32^\circ \quad \checkmark$

Part (b): Direction of friction, and the along-plane equation

The block is on the point of sliding down the plane, so friction, which always opposes impending motion, acts up the plane.

Resolving along the plane (taking “up the slope” as positive), three forces act in this direction: $H\cos32^\circ$ (up), the weight component $mg\sin32^\circ$ (down), and friction $\mu R$ (up, at its limiting value). For equilibrium: $H\cos32^\circ + \mu R = mg\sin32^\circ$

Part (c): Solving for $H$

Substitute the expression for $R$ from part (a) into the equation from part (b): $H\cos32^\circ + \mu\left(mg\cos32^\circ + H\sin32^\circ\right) = mg\sin32^\circ$ $H\cos32^\circ + \mu H\sin32^\circ = mg\sin32^\circ - \mu\, mg\cos32^\circ$ $H\left(\cos32^\circ + \mu\sin32^\circ\right) = mg\left(\sin32^\circ - \mu\cos32^\circ\right)$

Using $mg = 50$, $\mu = 0.3$, $\sin32^\circ \approx 0.5299$, $\cos32^\circ \approx 0.8480$: $\sin32^\circ - \mu\cos32^\circ \approx 0.5299 - 0.3(0.8480) \approx 0.2755$ $\cos32^\circ + \mu\sin32^\circ \approx 0.8480 + 0.3(0.5299) \approx 1.0070$

$H = \frac{50 \times 0.2755\ldots}{1.0070\ldots} \approx \frac{13.775}{1.0070} \approx 13.7\text{ N}$

Part (d): Newton’s third law

From part (a), the normal reaction on the block from the plane is: $R = 50\cos32^\circ + 13.68\sin32^\circ \approx 42.40 + 7.25 \approx 49.7\text{ N}$

By Newton’s third law, the force the block exerts on the plane is equal in magnitude and opposite in direction to $R$: a force of magnitude $\approx 49.7\text{ N}$, perpendicular to the plane, directed into the surface (whereas $R$ itself, the force of the plane on the block, points away from the surface). Note this is a genuinely different pair from “weight and normal reaction”. Those two forces both act on the block itself and are not a Newton’s third law pair.

Final answers

• (a) $R = mg\cos32^\circ + H\sin32^\circ$ (shown)
• (b) Friction acts up the plane; $H\cos32^\circ + \mu R = mg\sin32^\circ$
• (c) $H \approx \boxed{13.7}\text{ N}$
• (d) The block pushes on the plane with a force of $\approx \boxed{49.7}\text{ N}$, perpendicular to the plane, directed into the surface