Question

A beam is placed evenly on a pivot point. On one side a 10 N weight is placed 2 m from the pivot point and a 40 N weight a further 4 m from the pivot point. How far from the pivot point must an 80 N weight be placed to perfectly balance the beam horizontally?

Answer 1

Since , The beam is in equilibrium,

So,

anti dec. moment = dec. moment

=> 80x = (16 × 2) + 40 × (2 + 4)

=> 80x = 20 + 240

=> 80x = 260

=> x = 3.25 m

I.e, the weight must be placed in 3.25 m far from pivot point