Question

Two blocks A and B of mass $m_A=10kg$ and $m_B=20kg$ are place on rough horizontal surface. The blocks are connected with a string. If the coefficient of friction between block A and ground is ${\mu }_A=0.9$ and between block B and ground is ${\mu }_B=0.3$, find the tension in the string in situation as shown in Fig. Forces 120N and 100N start acting when system is at rest?

Answer 1

Let us that the system moves towards left. Then as it is clear from FBD,

$F_{driving}$= 120 -100 = 20N

$f_{resisting}={\mu }_A{N}_A+{\mu }_B{N}_B$

= 90 + 60 = 150 N

As $F_{resiststing}>F_{driving}$, therefore, it can be concluded that the system is stationary. Now there may be two possibilities.

The friction between both blocks and ground should be static.

The friction between one block and ground is static and between the other block and ground is limiting.

If friction between both blocks and ground is static, the tension in string should be zero.

The fraction on block A, $f_A$ = 120N

The fraction on block B. $f_B$ = 100N

But $(f_A{)}_{max}$ = 90N and $(f_B{)}_{max}=60N$

Hence, the friction on both blocks cannot be static.

Hence, friction on one block is static and other block is static and on other block should be limiting.

Assuming that the 10 kg block reaches limiting friction first then using FBD's.

If block A is at rest

120 = T + 90 = T = 30N

From FBD of B: T+f = 100

$\therefore$ 30 +f = 100

Thus, f= 70N which is not possible as the limiting value is 60 N for this surface of block.

Therefore, out assuming is wrong and now taking the 20-kg surface to be limiting, we have

From FBD of B,

T + 60 = 100 = T= 40N

From FBD of A,

Also f+T = 120 =f = 80N

This is acceptable as the static friction at this surface should be less than 90N.

Hence, the tension in the string is T = 40N.