Term 1 review: moments, parallel forces, equilibrium and couples
Answers are for parents and teachers.
1
A force (8, −6) N acts at A(−2, 5) m. Find its moment about the origin (anticlockwise positive) and the arm.
M = (−2)(−6) − 5 × 8 = 12 − 40 = −28 N·m (clockwise). |F| = 10 N, so L = 28 / 10 = 2.8 m.
2
Parallel forces of 21 N at A and 35 N at B act on a light rod with AB = 60 cm. Find the resultant and where it acts (distance from A) (a) when both point the same way, (b) when they point opposite ways (say which way R points).
(a) R = 56 N, 21 × AC = 35 × (60 − AC), so 56 AC = 2100 and AC = 37.5 cm. (b) R = 14 N in the direction of the 35 N force, acting beyond B: 21 × (60 + x) = 35 × x, so 1260 = 14x, x = 90 cm beyond B (150 cm from A).
3
A uniform plank 3 m long, mass 20 kg (g = 10), rests on a support at its end A and a support at C, 2.5 m from A. A person of mass 30 kg stands at the other end B. Find the upward reactions at A and C.
Weights: 200 N at 1.5 m, 300 N at 3 m. About A: 2.5 RC = 200 × 1.5 + 300 × 3 = 1200, so RC = 480 N. RA = 500 − 480 = 20 N.
4
A uniform ladder 5 m long, weight 200 N, stands with its foot 3 m from a smooth wall (top 4 m up) on smooth ground, held by a horizontal rope at its foot. A man of weight 400 N climbs it. The rope breaks at 225 N. How far along the ladder can he go?
T = S. About the foot: 4S = 200 × 1.5 + 400 × 3f = 300 + 1200 f. With S = 225: 900 = 300 + 1200 f, so f = 0.5 and he can climb 2.5 m along the ladder.
5
Square ABCD of side 0.5 m has forces of 16 N from A to B and from C to D, forces of 10 N from B to C and from D to A, and a further couple of 9 N forces with arm 0.4 m clockwise. Find the resultant couple (anticlockwise positive).
The 16 N pair is anticlockwise: +16 × 0.5 = +8. The 10 N pair (B to C up on the right, D to A down on the left) is anticlockwise: +10 × 0.5 = +5. The extra couple: −9 × 0.4 = −3.6. Total: +9.4 N·m, anticlockwise.