Question

Consider the given 3-block system.

Friction coefficient (static as well as kinetic) between surfaces are $\mu_1, \mu_2, \mu_3$, as shown. A horizontal force F of constant magnitude is to be applied on any of the block. Choose the correct options regarding motion of blocks.

• A. If force F is applied on block S maximum possible value of F for which there will be no slipping between any blocks is 30 N.
• B. If force F is applied on block S, maximum value of friction force acting between R and P will be 10N.
• C. If force F is applied on block P, maximum value of F for which there will be no slipping between any blocks is 15 N.
• D. If force F is applied on block P, maximum value of friction force acting between S and R will be 20 N.

Answer 1

Answer: (A), (D)

Options A and D

Answer 2

• When F is applied on S and ground friction is neglected, the strongest limitation comes from the weakest interface between S–R at $\mu_2=0.2$.
• Normal at S–R = $(m_R + m_P)g =10g$, so $f_{2,\max}=0.2\times10g=19.6\approx20$ N.
• System mass =15 kg, maximum acceleration $a_{\max}=f_{2,\max}/(m_R+m_P)=19.6/10=1.96$ m/s², so $F_{\max}=15\times1.96\approx30$ N.

• Maximum friction at R–P interface is $f_{3,\max}=0.3\times5g=14.7$ N, but the actual friction needed is only $5a=5\times1.96=9.8$ N. Hence option B is false.

• When F is applied on P, the limiting interface is again S–R giving $f_{2,\max}=19.6$ N, so option D is true. Option C (15 N) is far below the true limit (~30 N).