A crate of mass 25 kg rests in equilibrium on rough horizontal ground. A rope attached to the crate is pulled with a constant tension of 80 N, the rope making an angle of 30° with the horizontal. The coefficient of friction between the crate and the ground is 0.4. Take g = 10 m…

Answer: ((a)) horizontal component ≈ 69.3 N; vertical component = 40 N ((b)) R = 210 N ((c)) friction ≈ 69.3 N, acting horizontally opposite to the pull; since this is less than the limiting friction of 84 N, equilibrium is possible. Setting up the forces Four forces act on the crate: its weight W (down), the normal reaction R (up), the tension T = 80 N in the rope (at 30° above the horizontal), and the friction force F (horizontal, opposing the crate's tendency to sli…

A Level Maths 9709: Equilibrium With an Angled Pull and Friction

A crate of mass $25\text{ kg}$ rests in equilibrium on rough horizontal ground. A rope attached to the crate is pulled with a constant tension of $80\text{ N}$ , the rope making an angle of $30^\circ$ with the horizontal. The coefficient of friction between the crate and the ground is $0.4$ . Take $g = 10\text{ m s}^{-2}$ .

(a) Find the horizontal and vertical components of the $80\text{ N}$ tension. [2]

(b) By resolving forces perpendicular to the ground, find the normal reaction $R$ between the crate and the ground. [2]

(c) By resolving forces horizontally, find the frictional force acting on the crate, and state its direction. Hence show that the crate can indeed remain in equilibrium. [3]

Setting up the forces

Four forces act on the crate: its weight $W$ (down), the normal reaction $R$ (up), the tension $T = 80\text{ N}$ in the rope (at $30^\circ$ above the horizontal), and the friction force $F$ (horizontal, opposing the crate’s tendency to slide in the direction the rope is pulling).

The weight of the crate is: $W = mg = 25 \times 10 = 250\text{ N}$

Part (a): Components of the tension

Resolving the $80\text{ N}$ tension into horizontal and vertical components: $T_x = 80\cos30^\circ = 80 \times 0.8660\ldots \approx 69.3\text{ N}$ $T_y = 80\sin30^\circ = 80 \times 0.5 = 40\text{ N}$

Part (b): Normal reaction $R$

Resolving perpendicular to the ground (taking “up” as positive), the crate does not accelerate vertically, so the forces balance: $R + T_y - W = 0$ $R = W - T_y = 250 - 40 = 210\text{ N}$

The upward pull of the rope reduces the normal reaction below the crate’s weight.

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 tension: $F = T_x \approx 69.3\text{ N}$

Since the tension pulls the crate horizontally in one direction, friction acts horizontally in the opposite direction, resisting the crate’s tendency to slide.

To check this is physically possible, compare $F$ with the maximum available friction, $\mu R$ : $\mu R = 0.4 \times 210 = 84\text{ N}$

Since the required friction ( $\approx 69.3\text{ N}$ ) is less than the maximum available friction ( $84\text{ N}$ ), there is enough friction available (with a margin of about $14.7\text{ N}$ to spare) for the crate to remain in equilibrium, exactly as stated in the question.

Final answers

(a) Horizontal component $\approx \boxed{69.3}\text{ N}$ ; vertical component $= \boxed{40}\text{ N}$
(b) $R = \boxed{210}\text{ N}$
(c) Friction $\approx \boxed{69.3}\text{ N}$ , acting horizontally opposite to the pull; since $69.3 < 84$ , the crate remains in equilibrium