A block of mass 10 kg is suspended at a distance of 20 cm from the centre of a uniform bar 1 m long. What force is required to balance it at its centre of gravity by applying the force at the other end of the bar? Ans. (40 N)


Given:  

Mass of the block  = m =  10 Kg
thus Magnitude of the First force = F₁ =  weight of the block = mg = 10 Kg ï½˜ 10 m s⁻² = 100 N 
First moment arm = L₁ =   20 cm = 0.2 m 
Second moment arm = L₂ = 0.5 m

To Find:

The magnitude of the second  Force  = F₂ =? N

Solution:   

According to the second condition of the equilibrium 

Σ T= 0 

or 

Clockwise Torque = Anticlockwise Torque

F₂ ï½˜ L₂= F₁ ï½˜ L₁

or

F₂ = `\frac {F₁ ｘ L₁}{L₂ }`

by putting values

F₂ = `\frac {100 N ｘ 0.2 m}{0.5 m }`

F₂ = 40 N

Thus the second force of 40 N is required to balance the centre of gravity at the other end of the bar

************************************

************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2020-21 Academic Skills and Knowledge (ASK)     

Note:  Write me in the comments box below for any query and also Share this information with your class-fellows and friends.