a force of 200N is applied downwards at one end of a bar, 5m long in total. if the bar is pivoted 1m from the?

Question

end where this force is applied, what force needs to be applied at the other end to keep the bar balanced?

Answer 1

Sum of clockwise moments = sum of anti-clockwise moments
So, 200 x 1 = F x 4
F = 50N

Hope this helps=)

Answer 2

Use the law of moments
F1*L1 = F2*L2
200*1m = F2*4m
F2 = 200/4m = 50N