Define a rigid body. 2. Distinguish between torque and couple. 3. A body of mass 3kg is pulled by a horizontal force F until the string makes 30° with vertical. Find F and tension T. (g=10m/s²).

The forces acting on the body are:

• Weight ($W$) acting vertically downwards.
• Tension ($T$) in the string acting along the string.
• Horizontal force ($F$) acting horizontally.

The string makes an angle of $30^\circ$ with the vertical.
The weight of the body is $W = mg$.

$$W = 3 \text{ kg} \times 10 \text{ m/s}^2 = 30 \text{ N}$$

For equilibrium, the net force in both horizontal and vertical directions must be zero.

Sum of vertical forces:

$$T \cos(30^\circ) - W = 0$$ $$T \cos(30^\circ) = W$$ $$T \cos(30^\circ) = 30 \text{ N}$$

Sum of horizontal forces:

$$F - T \sin(30^\circ) = 0$$ $$F = T \sin(30^\circ)$$

From the vertical forces equation:

$$T = \frac{30 \text{ N}}{\cos(30^\circ)}$$ $$T = \frac{30 \text{ N}}{\frac{\sqrt{3}}{2}}$$ $$T = \frac{60}{\sqrt{3}} \text{ N}$$ $$T = \frac{60\sqrt{3}}{3} \text{ N}$$ $$T = 20\sqrt{3} \text{ N}$$

Using $\sqrt{3} \approx 1.732$:

$$T \approx 20 \times 1.732 \text{ N}$$ $$T \approx 34.64 \text{ N}$$

From the horizontal forces equation:

$$F = T \sin(30^\circ)$$ $$F = (20\sqrt{3} \text{ N}) \times \frac{1}{2}$$ $$F = 10\sqrt{3} \text{ N}$$

Using $\sqrt{3} \approx 1.732$:

$$F \approx 10 \times 1.732 \text{ N}$$ $$F \approx 17.32 \text{ N}$$

The tension in the string is $\boxed{\text{34.64 N}}$ and the horizontal force is $\boxed{\text{17.32 N}}$.

4. Distinguish between stable and unstable equilibrium. (4 marks)

• Stable equilibrium occurs when a body, if slightly displaced from its equilibrium position, tends to return to that position. This happens when the center of gravity is at its lowest possible position, and any displacement raises it. An example is a cone resting on its base.
• Unstable equilibrium occurs when a body, if slightly displaced from its equilibrium position, tends to move further away from that position. This happens when the center of gravity is at its highest possible position, and any displacement lowers it. An example is a cone balanced on its tip.

5. A small object of mass 20kg is placed on a horizontal table. What is the magnitude of the normal reaction of the table on the body? (Take g = 10m/s²) (3 marks)

The object is on a horizontal table, so the forces acting on it are:

• Weight ($W$) acting vertically downwards.
• Normal reaction ($N$) from the table acting vertically upwards.

Since the object is at rest on the table, it is in vertical equilibrium. Therefore, the sum of vertical forces is zero.

$$N - W = 0$$ $$N = W$$

$$W = mg$$

Given $m = 20 \text{ kg}$ and $g = 10 \text{ m/s}^2$.

$$W = 20 \text{ kg} \times 10 \text{ m/s}^2$$ $$W = 200 \text{ N}$$

Since $N = W$:

$$N = 200 \text{ N}$$

The magnitude of the normal reaction of the table on the body is $\boxed{\text{200 N}}$.