Non-uniform boards, loaded rods and a bar on a wedge
Answers are for parents and teachers.
1
A non-uniform board AB, 6 m long, rests on supports C and D with AC = 1 m and AD = 4 m. A man of weight 300 N can stand 4.8 m from A before the board tips about D, and 0.6 m from A (the other way) before it tips about C. Find the weight of the board and the distance of its line of action from A.
About D: 300 × 0.8 = w(4 − x), so 240 = w(4 − x). About C: 300 × 0.4 = w(x − 1), so 120 = w(x − 1). Adding: 360 = 3w so w = 120 N, and x − 1 = 1, so x = 2 m.
2
Up and anticlockwise are positive. A uniform rod AB, 120 cm long and weighing 60 N, rests on a support 40 cm from A and is held level by a light vertical string at B. Find the tension and the reaction of the support. What weight hung at A makes the string slack?
About the support: the weight is 20 cm and B is 80 cm to the right: 80T = 20 × 60, so T = 15 N and R = 45 N. Slack: 40w = 20 × 60, so w = 30 N.
3
A uniform rod AB, 60 cm long and weighing 30 N, hangs level from vertical strings at A and at C, with AC = x cm. A load of 30 N hangs 15 cm from A. Each string can bear at most 40 N. Find the range of x. Up and anticlockwise are positive.
About A: xT_C = 30 × 30 + 30 × 15 = 1350, so TC = 1350 / x and TA = 60 − 1350 / x. TC ≤ 40 gives x ≥ 33.75; TA ≤ 40 gives x ≤ 67.5, true on the rod. So 33.75 ≤ x ≤ 60.
4
Two workers carry a load of 540 N hung from a light pole 1.8 m long (they hold its two ends A and B). The load hangs x m from A. B can carry at most 360 N. Find the greatest x and A’s load then. Up and anticlockwise are positive.
About A: 1.8 RB = 540x, so RB = 300x. 300x ≤ 360 gives x ≤ 1.2 m, and then RA = 540 − 360 = 180 N.
5
A uniform bar AB, 3 m long and weighing 60 N, rests with A on rough ground and a point C on a smooth wedge 1.2 m high. It is about to slip at an angle θ with sin θ = 4/5 (cos θ = 3/5). Find the reaction of the wedge and the coefficient of friction.
AC = 1.2 / 0.8 = 1.5 m. About A: 1.5R₂ = 60 × 1.5 × 0.6 = 54, so R₂ = 36 N. F = 36 × 0.8 = 28.8 N, N = 60 − 36 × 0.6 = 38.4 N, μ = 28.8 / 38.4 = 0.75.