Forces, Movement, Shape, and Momentum
6 marks · NormalA 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.
Calculate the weight of the beam and the weight of the box. Give your answers in newtons (N). (Take .)
Find the total downward force acting on the supports.
Calculate the upward force on the support that is closest to the box. Show your working.
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.
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(a) Weight of beam = 80 N, weight of box = 120 N
(b) 200 N
(c) Upward force on closest support = 152 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.