Question

A solid cube of mass 5 kg is placed on a rough horizontal surface, in xy-plane as shown. The friction coefficient between the surface and the cube is 0.4. An external force $\overrightarrow{F} = (3\hat{i}+4\hat{j}+20\hat{k})$ N is applied on the cube. (use g = 10 m/s²)

• A. The block starts slipping over the surface
• B. The friction force on the cube by the surface is 5 N
• C. The friction force acts in xy-plane at angle 127° with the positive x-axis in clockwise direction
• D. The contact force exerted by the surface on the cube is $\sqrt{925}$ N

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

1. Calculate Normal Force: The weight of the cube is $W = mg = 5 \text{ kg} \times 10 \text{ m/s}^2 = 50 \text{ N}$ (downwards). The applied force has a vertical component $F_z = 20 \text{ N}$ (upwards). Since the cube is on a horizontal surface and not accelerating vertically, the normal force $N$ is given by $N + F_z = W$. Thus, $N + 20 \text{ N} = 50 \text{ N}$, which gives $N = 30 \text{ N}$.

2. Calculate Applied Horizontal Force: The applied force is $\overrightarrow{F} = (3\hat{i}+4\hat{j}+20\hat{k})$ N. The horizontal component of the applied force is $\overrightarrow{F}_{xy} = 3\hat{i}+4\hat{j}$ N. The magnitude of this horizontal force is $F_{xy} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 \text{ N}$.

3. Calculate Maximum Static Friction: The maximum static friction force is $f_{s,max} = \mu_s N$, where $\mu_s = 0.4$ is the coefficient of static friction. So, $f_{s,max} = 0.4 \times 30 \text{ N} = 12 \text{ N}$.

4. Check for Slipping: Since the applied horizontal force $F_{xy} = 5 \text{ N}$ is less than the maximum static friction force $f_{s,max} = 12 \text{ N}$, the block does not start slipping. Thus, option A is incorrect.

5. Determine Friction Force: As the block does not slip, the static friction force is equal in magnitude and opposite in direction to the applied horizontal force. The friction force is $\overrightarrow{f} = -\overrightarrow{F}_{xy} = -(3\hat{i}+4\hat{j}) = -3\hat{i}-4\hat{j}$ N. The magnitude of the friction force is $|\overrightarrow{f}| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5 \text{ N}$. Therefore, option B is correct.

6. Determine Direction of Friction Force: The friction force vector is $\overrightarrow{f} = -3\hat{i}-4\hat{j}$. This vector lies in the xy-plane. To find its angle with the positive x-axis, we can consider its components. Let $\theta$ be the angle measured counter-clockwise from the positive x-axis. Then $\tan\theta = \frac{-4}{-3} = \frac{4}{3}$. Since both components are negative, the vector is in the third quadrant. The principal angle is $\arctan(4/3) \approx 53.13^\circ$. The angle in the third quadrant is $\theta \approx 180^\circ + 53.13^\circ = 233.13^\circ$. The angle measured clockwise from the positive x-axis is $360^\circ - 233.13^\circ = 126.87^\circ$. This is approximately $127^\circ$. Therefore, option C is correct.

7. Calculate Contact Force: The contact force exerted by the surface on the cube is the vector sum of the normal force and the friction force. The normal force is $\overrightarrow{N} = 30\hat{k}$ N (acting upwards). The friction force is $\overrightarrow{f} = -3\hat{i}-4\hat{j}$ N. The contact force is $\overrightarrow{F}_{contact} = \overrightarrow{N} + \overrightarrow{f} = 30\hat{k} + (-3\hat{i}-4\hat{j}) = -3\hat{i}-4\hat{j}+30\hat{k}$ N. The magnitude of the contact force is $|\overrightarrow{F}_{contact}| = \sqrt{(-3)^2 + (-4)^2 + (30)^2} = \sqrt{9 + 16 + 900} = \sqrt{25 + 900} = \sqrt{925}$ N. Therefore, option D is correct.