A uniform rod 1m long with a weight of 6 N can be supported in a horizontal position on a sharp edge with weights of 10 N and 15 N suspended from its ends. What is the position of the point of balance? (Ans. 0.41 m)


Given:

Length of the rod = 1 m 
Weight of the rod Fr = 6N 
Weight F₁ = 10 N 
Weight F₂ = 15 N 
Moment arm of Rod = r = x
Moment arm of Weight F₁ = r₁  = 50 cm + x 
Moment arm of Weight F₂ = r₂  = 50 cm - x 
 

To Find:

Position of point of balance = x =  ?

Solution:

Taking point P as the axis of rotation and applying 2 conditions of equilibrium that is

Σ T = 0

or

Σ Anti-clockwise torques =  Σ Clockwise torques 

r Fr  + r₁ F₁ = r₂ F₂

x ｘ 6 + (50 + x) 10 = (50 - x) 15 

6x +500 +10 x = 750 - 15 x

500 + 16 x = 750 - 15 x

16 x + 15 x = 750 - 500 

31x = 250

x = `\ frac {250}{31}`

x = 8.06 cm

Thus the location balance point from Weight F₁ = r₁  = 50 + x = 50 cm + 8.06 cm = 58.06 cm = 0.580 m

and


the location balance point from Weight F₂ = r₂  = 50 cm - x = 50 cm - 8.06 cm = 41.94 cm = 0.419 m


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