Problem 9: A meter rule is supported on a knife edge placed at the 40 cm graduation. It is found that the meter rule balances horizontally when a mass which has a weight of 0.45 N is suspended at the 15 cm graduation. As shown in the diagram.
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num34.png)
(a) Calculate
the moment about the knife edge in this balanced condition of the force due to
the mass of the rule.
(b) If weight
of the ruler is 0.90 N then find the position of its center o f gravity.
Solution
Given data
- Distance of the weight W1to the nearer
end A = 15 cm = 0.15 m - Distance of the knife edge to the nearer end A = 40
cm = 0.40 m - Distance between the weight and knife edge = 40 – 15
= 25 cm = 0.25 m - Weight W1 = 0.45 N
- Weight of rule =W = 0.9 N
Required
- Torque of W1 about the knife edge =?
- Position of the CG =?
Now let, Torque of W1 about the knife edge = τ
Position of the CG = C
(a) Now to find τ,
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num35.png)
(b) To find the position of the center of gravity,
we apply the conditions of equilibrium.
Since there is no force acting along the
x-direction, therefore, ∑Fx = 0
For the forces in vertical directions, we must have, ∑Fy = 0
OR
R – W1 – W = 0
(Please note that the forces acting in the upward
direction are taken to be positive and those downward are taken as negative).
Putting the values, we have
R – 0.45 – 0.9 = 0
OR R – 1.35 = 0
OR R = 1.35 N
Now apply second condition of
equilibrium, ∑τ = 0 letting A to be the point of rotation.
![](https://mashalscienceacademy.com/wp-content/uploads/2021/03/c2p11num36.png)
So the center of gravity would be
at a distance of 52.5 cm from the end A of the rule.
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