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Question 2: [Turning effect of forces]

The diagram shows a plank of mass 5.0 kg and length 3.00 m resting horizontally on two trestles, P and Q, which are a distance of 2.50 m apart. When a student of mass 60.0 kg walks along the plank from one trestle to the other, the plank sags.

(a) Explain why the sag increases as the student walks towards the middle of the plank.

(b) Calculate the downward force exerted on each trestle when the student is at a distance of

(i) 0.50 m from trestle P,

(ii) 1.25 m from trestle P.

(g = 10.0 N/kg)

Reference: Past Exam Paper – N91 / P2 / QA2

Solution 2:

(a)

Assuming the plank is uniform, the weight of the plank acts at the midpoint of the plank causing the sag. As the bog walks from P towards the middle of the plank, the combined centre of gravity of the system of plank and boy acts in between the boy and the midpoint of the plank causing a larger sag. As the boy moves even nearer to the midpoint, the sag increases as the combined c.g. moves towards the centre of the plank.

(b)

(i) When x = 0.50 m from P,

By the Principle of Moments, moment about P

R_Q ×2.50 = (600×0.50) + (50×1.25)

R_Q = 145 N

For vertical equilibrium,

R_P + R_Q = 600 + 50 = 650 N

R_P = 650 – R_Q = 650 – 145 = 505 N

(ii) When x = 1.25 m from P, both the forces 600 N and 50 N act at the centre of mass of the plank.

By symmetry, R_P = R_Q            where  R_P + R_Q = 600 + 50 = 650 N

R_P = R_Q = 650/2 = 325 N