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).

SECTION B 1. Define a rigid body. (2 marks) A rigid body is an idealization of a solid body in which deformation is neglected. In other words, the distance between any two given points of a rigid body remains constant in time regardless of external forces or torques exerted on it. 2. Distinguish between torque and couple. (4 marks) Torque* is the rotational equivalent of linear force. It is a measure of the force that can cause an object to rotate about an axis. Torque is calculated as the product of the force and the perpendicular distance from the pivot to the line of action of the force ( = rF). A couple* is a pair of forces, equal in magnitude, opposite in direction, and separated by a perpendicular distance, that tends to produce rotation only. The net force of a couple is zero, but it produces a net torque (moment) about any point. 3. A body of mass 3kg is suspended by an inextensible string from a nail O and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 30°. Calculate the value of F and the tension in string. (Take g = 10m/s²). (7 marks) Step 1: Draw a free-body diagram and resolve forces. 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^ with the vertical. The weight of the body is W = mg. W = 3 kg × 10 m/s^2 = 30 N Step 2: Apply equilibrium conditions. For equilibrium, the net force in both horizontal and vertical directions must be zero. Sum of vertical forces: T (30^) - W = 0 T (30^) = W T (30^) = 30 N Sum of horizontal forces: F - T (30^) = 0 F = T (30^) Step 3: Calculate the tension (T). From the vertical forces equation: T = 30 N(30^) T = 30 Nsqrt(3)2 T = (60)/(sqrt(3)) N T = 60sqrt(3)3 N T = 20sqrt(3) N Using sqrt(3) ≈ 1.732: T ≈ 20 × 1.732 N T ≈ 34.64 N Step 4: Calculate the horizontal force (F). From the horizontal forces equation: F = T (30^) F = (20sqrt(3) N) × (1)/(2) F = 10sqrt(3) N Using sqrt(3) ≈ 1.732: F ≈ 10 × 1.732 N F ≈ 17.32 N The tension in the string is 34.64 N and the horizontal force is 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) Step 1: Identify the forces acting on the object. 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. Step 2: Apply equilibrium condition. 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 Step 3: Calculate the weight (W). W = mg Given m = 20 kg and g = 10 m/s^2. W = 20 kg × 10 m/s^2 W = 200 N Step 4: Determine the normal reaction (N). Since N = W: N = 200 N The magnitude of the normal reaction of the table on the body is 200 N.