Question

a spring balance is kept horizontally it's too in Sah for pulled by a force of 5 kg then what will be its reading

Answer 1

If a block is attached to a spring which has 1 end fixed to the wall then the force which is required to shift the block by ‘x’ distance is given by the following formula: 

F = - kx 

Where k is the spring constant and x is the displacement.

Now in the present situation each block of 5kgwt is attached to the spring on each side. Force by which the horizontal spring is pulled at each end is equal to 5 kgwt or = 5N. Since the vertical forces of Normal reaction and force of Gravitation balance each other, the horizontal force exerted on each side is also of 5N and are opposite to each other. 

Now for the horizontally placed spring balance, let the displacement towards right due to 5 kgwt be ‘x’. since the displacement for the other 5kgwt is towards left , hence its displacement will be ‘-x’. 

Now let F1 be the force exerted by the weight on right. Its value will be equal to: 

F1 = -kx.

Let F2 be the force exerted by weight on left. Its value will be equal to : 

F2 = -k(-x) ………..(as the displacement is in the opposite direction)

Now F1 and F2 are equal and opposite to each other, Hence net force will be 

F2 – F1……………...(force is a vector quantity) 

Net force= -k(-x) – (-kx) 

= +kx +kx 

= +2kx 

Hence the spring balance will show a reading of 5kg +5kg = 10kg. Which is the required answer.