The two conditions of equilibrium: boards, rods and strings
Answers are for parents and teachers.
1
Up and anticlockwise are positive. A light rod AB, 90 cm long, rests on supports at A and B and carries 45 N at 30 cm from A and 15 N at 60 cm from A. Find RA and RB.
About B: 90 RA = 45 × 60 + 15 × 30 = 3150, so RA = 35 N and RB = 60 − 35 = 25 N.
2
Take g = 10. A uniform plank AB of mass 25 kg and length 4 m rests on supports at its ends. A girl of mass 40 kg stands 1 m from B. Up and anticlockwise are positive. Find the reactions in newtons.
Weights 250 N (at 2 m from A) and 400 N (at 3 m from A). About A: 4 RB = 250 × 2 + 400 × 3 = 1700, so RB = 425 N and RA = 650 − 425 = 225 N.
3
A uniform rod AB, 1.5 m long and weighing 30 N, hangs level from vertical strings at A and B. A weight of 60 N hangs x m from A. Find x if TA = 2 TB. Up and anticlockwise are positive.
Total 90 N, so TB = 30 N and TA = 60 N. About A: 1.5 × 30 − 30 × 0.75 − 60x = 0, so 45 − 22.5 = 60x and x = 0.375 m (37.5 cm).
4
A uniform plank AB, 6 m long and weighing 360 N, rests on a support at A and a support at C, where C is 1 m from B. A box of weight 240 N stands on the plank x m from A. Find x if the two reactions are equal. Up and anticlockwise are positive.
Each reaction is (360 + 240) / 2 = 300 N. Moments about C (AC = 5 m): −5 × 300 + 2 × 360 + (5 − x) × 240 = 0, so (5 − x) × 240 = 780, 5 − x = 3.25 and x = 1.75 m.
5
A rod has 50 N up at A and 50 N down at a point 20 cm from A. Up and anticlockwise are positive. Find ΣY, then ΣM about A and about the other point. Is the rod in equilibrium?
ΣY = 0. About A: 20 × (−50) = −1000 N·cm. About the other point: (−20)(50) = −1000 N·cm. The same clockwise moment about every point: the rod is not in equilibrium, it turns (a couple).