**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.

(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 W
_{1}to 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 W
_{1}= 0.45 N - Weight of rule =W = 0.9 N

**Required**

- Torque of W
_{1}about the knife edge =? - Position of the CG =?

Now let, Torque of W_{1} about the knife edge = τ

Position of the CG = C

(a) Now to find τ,

(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, ∑F_{x} = 0

For the forces in vertical directions, we must have, ∑F_{y} = 0

OR

R – W_{1} – 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.

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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