Forces, Movement, Shape, and Momentum

6 marks · Normal

A 2.0 m long uniform beam of mass 8.0 kg is supported at each end. A box of mass 12 kg is placed 1.2 m from one end of the beam.

(a)

Calculate the weight of the beam and the weight of the box. Give your answers in newtons (N). (Take .)

[2 marks]

(b)

Find the total downward force acting on the supports.

[1 mark]

(c)

Calculate the upward force on the support that is closest to the box. Show your working.

[3 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) Weight of beam = 80 N, weight of box = 120 N

(b) 200 N

Common mistakes: Not including mass of the beam in total weight; Wrong pivot position in moments

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Correctly use W = m × g for beam and box.

A1 (a) · 1 mark · Correct numerical answers for both weights (80 N, 120 N).

B1 (b) · 1 mark · Correctly sum weights for total downward force (200 N).

M2 (c) · 1 mark · Set up moment equation about other support using correct distances and weights. (follow-through)

A2 (c) · 1 mark · Correct value for moment for each force. (follow-through)

A3 (c) · 1 mark · Correct upward force on support to within tolerance (152 N). (follow-through)

Worked solution

a) Weight of beam = . Weight of box = .

b) Total downward force = weight of beam + weight of box = .

c) Take moments about support furthest from the box: - Distance of box from that support = 2.0 m - 1.2 m = 0.8 m - Centre of gravity of beam is at 1.0 m from either end: distance from that support is 2.0 m - 1.0 m = 1.0 m Let F = upward force at closest support (to the box), so the other is (200 - F). Take moments about furthest support: \[ F \times 2.0 = (80 \times 1.0) + (120 \times 0.8)\implies F \times 2.0 = 80 + 96 = 176 \] \[ F = \frac{176}{2.0} = 88\ \] Error! (box is closer to this support, so try moments about the other end) Let's set F as the upward force at left support (closest to box, at 1.2 m from left), Take moments about the right support (distance from left = 2.0 m): - Box: distance from right end = 2.0 - 1.2 = 0.8 m - Beam's centre: 1 m from left, so from right = 2.0 - 1.0 = 1.0 m \[ (F)\times 2.0 = (120 \times 0.8) + (80 \times 1.0) \] \[ F = (96+80)/2.0 = 176/2.0 = 88 N \] But we want the force at the support _closest to box_. The net upward force is 200 N in total, so the upward force at the right support is 48 N, so at the left (closest to box): 200 N - 48 N = 152 N. So the upward force on the support closest to the box is 152 N.