Several parallel forces and their resultant
Answers are for parents and teachers.
1
Two parallel forces, 45 N and F, have a resultant of 75 N whose line is 10 cm from the 45 N force. Take the direction of R as positive. Find F and the distance between the lines when (a) the 45 N force points the same way as R, (b) the 45 N force points against R.
(a) F = 75 − 45 = 30 N; 45 × 10 = 30 × d gives d = 15 cm on the other side, so the forces are 25 cm apart. (b) F = 75 + 45 = 120 N; 450 = 120 × d gives d = 3.75 cm on the same side, so they are 10 − 3.75 = 6.25 cm apart.
2
Four like forces of 5 N, 10 N, 20 N and 15 N, all pointing up, act at A, B, C, D on a line with AB = 4 cm, BC = 6 cm, CD = 2 cm. Find the resultant and its distance from A.
R = 50 N up. Positions from A: 0, 4, 10, 12. 10 × 4 + 20 × 10 + 15 × 12 = 420, so xR = 420 / 50 = 8.4 cm from A.
3
Up is positive. The figure shows forces on a light rod: 20 N down at A, 25 N up at 4 cm from A, 10 N up at 9 cm from A and 10 N down at B, 15 cm from A. Find the resultant and where its line cuts the rod.
R = −20 + 25 + 10 − 10 = 5 N up. Moments about A: 4 × 25 + 9 × 10 + 15 × (−10) = 100 + 90 − 150 = 40, so xR = 40 / 5 = 8 cm from A.
4
Up is positive, lengths in cm. A light rod AB is 20 cm long. It carries K up at A, 40 N down at 6 cm from A, 90 N up at 14 cm from A and F down at B. The resultant is 50 N upwards and acts 12 cm from A. Find K and F.
Forces: K − 40 + 90 − F = 50, so K − F = 0. Moments about A: 6 × (−40) + 14 × 90 − 20F = 50 × 12, so −240 + 1260 − 20F = 600, 20F = 420 and F = 21 N. Then K = 21 N.
5
A girl says: ‘Two forces of 6 N up at A and 6 N down at B, 10 cm apart, have a resultant of 0 N, so they do nothing to the rod.’ Is she right? Find the sum of the moments about A and about B.
No. The resultant force is zero, but the rod still turns. Up positive, B to the right of A: about A the sum is 10 × (−6) = −60 N·cm; about B it is (−10)(6) = −60 N·cm. The same clockwise turning effect about any point: this is a couple (Unit 3).