A horizontal force of 500 N pulls two masses 10 kg and 20 kg (lying on a frictionless table) connected by a light string. What is the tension in the string ? Does the answer depend on which mass end the pull is applied?

Explanation

Here, `m_1 = 10 kg, m_2 = 20 kg and F = 500 N` Case (a) : When the force F is applied on mass `m_1`: Let `T_1` be the tension in the connecting string. The free body diagram with all the forces in the horizontal directions of the masses `m_1 and m_2` are shown in figure (a). Now, applying Newton.s second law of motion to masses m, and m, we get
` F-T_1 =m_1 a`
` and T_1 = m_2 a`
Eliminating a from eqns (i) and (ii) we get,
`(F-T_1)/(m_1 ) = (T_1)/(m_2) or T_1 =(m _2 F)/(m_1 +m_2) = ( 20 xx500 )/( 10 +20 ) = 333.3 N`
Case (b) : When the force is applied on mass my. With reference to figure (b), we can write,
` T_2 = m_1 a `
and ` F-T_2 = m_2 a`
Again eliminating a from eqns (iii) and (iv) we get,
`T_2=(m_1 F)/( m_1 +m_2) = (10 xx 500 )/(10 +20) = 166.7 N.`
So, the tensions in both the cases are different, i.e., it depends on which mass the force is applied. It may be noted that the acceleration is same in both the cases.