A uniform plank of weight 100 N and length 4.0 m rests horizontally on two supports. Support P is at the left end of the plank and support Q is at the right...

Marking Scheme

• (a) [1 mark]: For an object in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments [1]
• (b) [1 mark]: Total \(= 100 + 300 = 400\) N [1]
• (c) [3 marks]: Moments about P: \(R_Q \times 4.0 = 100 \times 2.0 + 300 \times 1.0\) [1]; \(R_Q \times 4.0 = 200 + 300 = 500\) [1]; \(R_Q = 125\) N [1]
• (d) [2 marks]: \(R_P + R_Q = 400\); \(R_P = 400 - 125\) [1]; \(= 275\) N [1]
• (e) [1 mark]: At the centre of the plank, 2.0 m from each support [1]

Explanation

Part (a): The principle of moments states that for a body in rotational equilibrium, the total clockwise moment about any pivot equals the total anticlockwise moment about that same pivot.

Part (b): The total downward force is the sum of the plank's weight and the crate's weight: \(100 + 300 = 400\) N. For vertical equilibrium, the total upward reaction (\(R_P + R_Q\)) must also equal 400 N.

Part (c) - Moments about P: Taking moments about P eliminates \(R_P\) (its moment arm is zero). The clockwise moments come from the plank's weight (acting at 2.0 m, the midpoint) and the crate's weight (acting at 1.0 m):

\(R_Q \times 4.0 = (100 \times 2.0) + (300 \times 1.0)\)

\(R_Q \times 4.0 = 200 + 300 = 500\)

\(R_Q = 125\) N

Part (d): From vertical equilibrium:

\(R_P = 400 - 125 = 275\) N

The support nearer to the crate (P) bears the larger share of the load, which makes physical sense.

Part (e): For the reactions to be equal (\(R_P = R_Q = 200\) N each), the combined centre of gravity of the plank-plus-crate system must be at the midpoint of the plank (2.0 m from each end). The plank's own weight already acts at 2.0 m. To keep the system balanced, the crate must also be placed at 2.0 m from P - the centre of the plank.