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.
M=Fd
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).
MA2 = FĂ—d = 1000NĂ—0.5m = 500 Nm
I could not really grasp how you determined this equation for the anti clockwise moment, could you please explain it further?
The bar has a weight of 1000 N, we consider weight as acting through the centre of the object (at the 1m mark of this 2m metal bar) downwards in the same direction as the added weights
As this point is to the left of the pivot we consider it as part of the forces moving around the pivot in the anti-clockwise direction ( in addition to the 200N at the end of the bar)
So we add it as part of the anti-clockwise moment combined with moment from the 200N and we have to balance the clockwise moment with both these values