Question

A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-hand end of the bar. What distance x from the left-hand end of the bar should an object with mass 0.900 kg be suspended so the bar is balanced in a horizontal position?

Answer 1

Explanation and answer:

This type of question can be clarified and sometimes solved by drawing a proper diagram or sketch.  (see below)

Solution:

Since we do not know the reaction of the support, we can take moments about the support (thereby eliminating its involvement).

CCW moment = 0.900(5.00/2 - x) kg-m

CW moment = 0.300*(5.00/2-2.00)) = 0.150 kg-m

At equilibrium, CCW moment = CW moment, so

0.900(2.50-x) = 0.150

Expand and solve

2.25 - 0.900x = 0.150

0.900x = 2.25-0.15 = 2.10

x = 2.10 / 0.900 = 2.33 m  (to nearest cm)