THE TURNING EFFECT OF FORCES
4 marks · NormalA uniform horizontal rod of length 1.8 m and weight 90 N is supported at its two ends, labelled P and Q. A load of 60 N is hung from the rod, 0.3 m from end P.
Taking moments about point P, calculate the upward force at Q.
State the upward force at P.
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) 38 N · must show (Upward force at Q) × 1.8 = 60 × 0.3 + 90 × 0.9, 38 N
(b) 112 N · must show (Total up) – (Upward force at Q), 112 N
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Sets up moments equation about P correctly.
M2 (a) · 1 mark · Substitutes correct distances and forces.
A1 (a) · 1 mark · Calculates upward force at Q as 38 N, with units.
A2 (b) · 1 mark · Calculates upward force at P as 112 N, with units. (follow-through)
Worked solution
a) Take moments about P (anticlockwise = clockwise):
Correction: The distance for the rod's weight from P is 0.9 m (rod's centre at 0.9 m). So:
But now, let's double-check the total upward force: Upward at P + Q = 60 + 90 = 150 N (all weights down).
So, final values: Q = 55 N, P = 95 N