Forces and Equilibrium: Question 3

Worked solution

Setting up the forces

The parcel has weight $W = mg = 8 \times 10 = 80\text{ N}$, acting vertically downwards. On the inclined plane, it is convenient to resolve perpendicular to the plane and along the plane (parallel to the line of greatest slope), since the normal reaction and (for part (a)) the tension act in these directions.

Resolving the weight into these two directions: $\text{perpendicular component} = mg\cos25^\circ, \qquad \text{parallel component} = mg\sin25^\circ$

Part (a): Normal reaction $R$

The tension acts along the plane (up the slope), so it has no component perpendicular to the plane. Only the weight’s perpendicular component and the normal reaction act in this direction, and since the parcel does not move off the plane, these balance: $R = mg\cos25^\circ = 80 \times \cos25^\circ = 80 \times 0.9063\ldots \approx 72.5\text{ N}$

Part (b): Direction of friction, and finding $T$

The parcel is on the point of sliding up the plane. Friction always opposes the direction of impending motion, so here friction acts down the plane.

Since the parcel is in limiting equilibrium, friction takes its maximum possible value: $F = \mu R = 0.2 \times 72.5\ldots \approx 14.5\text{ N}$

Resolving along the plane (taking “up the slope” as positive), three forces act along this direction: the tension $T$ (up), the weight component $mg\sin25^\circ$ (down), and the friction $F$ (down, since it opposes the impending slide up). For equilibrium: $T - mg\sin25^\circ - F = 0$ $T = mg\sin25^\circ + F$

Calculating $mg\sin25^\circ$: $mg\sin25^\circ = 80 \times \sin25^\circ = 80 \times 0.4226\ldots \approx 33.8\text{ N}$

So: $T = 33.8\ldots + 14.5\ldots \approx 48.3\text{ N}$

Final answers

• (a) $R \approx \boxed{72.5}\text{ N}$
• (b) Friction acts down the plane; $T \approx \boxed{48.3}\text{ N}$