IMAT 2011 Q71 [Forces]

A uniform bar of length 2.0 m and weight 1000 N has its centre of gravity at its centre. The bar is pivoted in the position shown, and supports a weight of 200 N in the position shown in the diagram.


What weight is needed at position P to balance the bar?

A. 600N
B. 800N
C. 1000N
D. 1600N
E. 1800N


For a uniform bar to be “stable” or in equilibrium, the Clockwise Moment must equate the Anti-Clockwise moment around the pivot.

A moment (M) in Newton meters is equal to the product of the Force (F) in Newtons and the perpendicular distance from the pivot (d) in meters.


With the provided values we are able to deduce the Anti-clockwise moment.

However beforehand, we must look at any additional information that may come into play.

Here, this information, or trick per say, is “A uniform bar of length 2.0 m and weight 1000 N has its centre of gravity at its centre”.

This means that for the Anti-clockwise moment, we must calculate 2 sets of values and add them to give us the final Anti-clockwise moment-

So it will look something like this:

M_{A1}= F \times d= 200N \times (2m - 0.5 m) = 200N \times 1.5 = 300 Nm

M_{A2} = F \times d = 1000N \times 0.5m = 500 Nm

Since the weight of the bar acts on it’s center of gravity at the centre of the bar (1m), the distance from the pivot will be 0.5m

M_{AT}= M_{A1} + M_{A2} = 300 Nm + 500 Nm = 800 Nm

Now, we proceed to look at the Clockwise Moment-

M_{C}= F \times d = 0.5F

We learn that we need to find this force or weight, F.

We can use the concept of Anti-clockwise moment= Clockwise Moment to find it:

M_{AT}= M_{C},

where M_{AT}= 800 Nm and M_{C}= 0.5F

So we can say

0.5F= 800 Nm

F= {\frac {800 Nm}{0.5 m}}= 1600 N

Which makes the weight needed at Position P to balance the bar 1600 N or choice (D).