Question

A uniform bar 5 meter weighs 500 N. A 50 N weight is hung at the end of the bar. At what point must the bar be supported to remain horizontal?

Answer 1

Answer:

0.25 metres from the centre on the same side of the weight.

Explanation:

Total length(d) = 5m

weight of the bar = 500N

weight of   50N is added on one end

According to the principle of moments of force (Sum of clockwise moments = sum of anticlockwise moments)

50 x 2.5 = 500 x distance from the centre

=> 125 = 500 x distance from the centre

=> distance from the centre = 125/500

=> distance from the centre = 0.25 m

Therefore, the support must be kept 0.25 metres away from the centre on the same side.