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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 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,

(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.

So the center of gravity would be
at a distance of 52.5 cm from the end A of the rule.

3 Comments

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