3. 5 m A D с B 0.5 m 1.3 m Figure 2 A beam ADCB has length 5 m. The beam lies on a horizontal step with the end A on the step and the end B projecting over the edge of the step. The edge of the step is at the point D where DB = 1.3 m, as shown in Figure 2. When a small boy of mass 30 kg stands on the beam at C, where CB = 0.5 m, the beam is on the point of tilting. The boy is modelled as a particle and the beam is modelled as a uniform rod. (a) Find the mass of the beam. (3) A block of mass X kg is now placed on the beam at A. The block is modelled as a particle. (b) Find the smallest value of X that will enable the boy to stand on the beam at B without the beam tilting. (3) (c) State how you have used the modelling assumption that the block is a particle in your calculations. (1)

The principle of moments states that the for a body in equilibrium, the sum of the clockwise moments is equal to the sum of the anticlockwise moments about a point.

Sum of Clockwise moments = Sum of Anticlockwise moments
Moment of a force = mass × perpendicular distance.

At the point where the uniform rod is about to tilt, it is at equilibrium.

Anticlockwise moments = 30 × (1.3 - 0.5) = 24

Clockwise moments = m × (2.5 - 1.3)

where m is the mass of the end

Clockwise moments = m × 1.2

1.2 × m = 24

m = 20 Kg

Therefore, the mass of the rod is 20 kg

Assuming the boy moves to B and a block of mass X is placed at:

Anticlockwise moments = 30 × 1.3 = 39

Clockwise moments = 20 × 1.2 + X × 3.7

Clockwise moments = 24 + 3.7X

39 = 24 + 3.7X

3.7X = 15

X = 4.054 kg

Therefore, the smallest value of X of the block that will enable the boy to stand on the beam at B without the beam tilting is 4.054 kg

Assuming that the block is a particle allowed the assumption that the block touches the rod at only a point.

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