Forces and Equilibrium: Question 8

Worked solution

Setting up the forces

Four forces act on the crate: its weight $W$ (down), the normal reaction $R$ (up), the push $P = 100\text{ N}$ (at $20^\circ$ below the horizontal), and friction $F$ (horizontal, opposing the crate’s tendency to slide in the direction it is being pushed).

The weight of the crate is: $W = mg = 30 \times 10 = 300\text{ N}$

Part (a): Components of the push

Resolving the $100\text{ N}$ push into horizontal and vertical components: $P_x = 100\cos20^\circ = 100 \times 0.9397\ldots \approx 94.0\text{ N}$ $P_y = 100\sin20^\circ = 100 \times 0.3420\ldots \approx 34.2\text{ N} \text{ (downward)}$

Part (b): Normal reaction $R$

Resolving perpendicular to the ground (taking “up” as positive), the crate does not accelerate vertically. Unlike a rope pulling upward, this push has a downward vertical component, so it adds to the weight pressing the crate into the ground: $R - W - P_y = 0$ $R = W + P_y = 300 + 34.2\ldots \approx 334\text{ N}$

Part (c): Frictional force, and checking equilibrium is possible

Resolving horizontally, the crate does not accelerate, so the friction $F$ must exactly balance the horizontal component of the push: $F = P_x \approx 94.0\text{ N}$

To check this is physically possible, compare $F$ with the maximum available friction, $\mu R$: $\mu R = 0.5 \times 334.2\ldots \approx 167\text{ N}$

Since the required friction ($\approx 94.0\text{ N}$) is less than the maximum available friction ($\approx 167\text{ N}$), there is enough friction available (with a margin of about $73\text{ N}$ to spare) for the crate to remain in equilibrium, exactly as stated in the question.

Final answers

• (a) Horizontal component $\approx \boxed{94.0}\text{ N}$; vertical component $\approx \boxed{34.2}\text{ N}$ (downward)
• (b) $R \approx \boxed{334}\text{ N}$
• (c) Required friction $\approx \boxed{94.0}\text{ N}$; maximum friction $\approx \boxed{167}\text{ N}$; since $94.0 < 167$, the crate remains in equilibrium