A force of 200N is applied downwards at 1 end of the bar,5m long in total.?

If the bar is pivated 1m from the end where this force is applied, what force needs to be applied at the other end to keep the bar balanced?

Answers:
sum of clockwise moments=sum of anticlockwise moments
200N*1m = F*4m
200Nm/4m=F

Force then should be 50N
if not taking into account the centre of gravity then its 50N applied downwards
I wish the Internet had been around when WE were given homework questions! Sincerely - An Old Codger.
if neglecting the weight of e bar
clockwise moment=anitclockwise moment
(200)(1) = F (4)
F= 50N
50 N is applied downwards at the other end of the bar

if weight of bar is taken into consideration
centre of gravity of the bar would be in the centre (2.5m from each end) , hence
clockise moment = anticlockwise moment
(200)(1) = (Weight of bar)(1.5) + F(4)

F is the dowward force applied at the other end to keep the bar balanced.

=)

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