Question

Two blocks A and B of mass $m_A$= 10 kg and $m_B$ = 20 kg are place on rough horizontal surface. The blocks are connected with a string. If the coefficient of friction between the 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 the figure below? Forces 120 N and 100 N start acting when the system is at rest.

• A. 30 N
• B. 40 N
• C. 0 N
• D. None of these

Answer 1

The correct option is B

40 N

(i) Let us assume that 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=150N

As $F_{resisting}>F_{driving}$

Therefore it can be concluded that the system is stationary.

Now there may be two possibilities.

(i) The friction between both blocks and ground should be static.

(ii) The friction between one block and ground is static and between other block and ground is limiting.

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

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

The friction on block B, $f_B$ = 100 N

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

Hence the friction on both cannot be static.

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

Assuming that the 10kg block reaches limiting friction first then using free body diagram's.

T + 60 = 100 $\Rightarrow$ T = 40N

From Free body diagram of A

Also f + T = 120 $\Rightarrow$ f = 80 N

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

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