Question

Find the least horizontal force P to start motion of any part of the system of the three blocks resting upon one another as shown in Fig. The weights of blocks are A=300 N, B = 100N, and C =200 N. Between A and B, the coefficient of friction is 0.3, between B and C is 0.2, and between C and the ground is 0.1.

• A. 60 N
• B. 90 N
• C. 80 N
• D. 70 N

Answer 1

The correct option is A 60 N

Given

$m_A=30kg$ ${\mu }_{AB}=0.3$

$m_B=10kg$ ${\mu }_{BC}=0.2$

$m_C=20kg$ ${\mu }_{CG}=0.1$

For slipping at $AB$ contact $F=300\times 0.3=90N$

For slipping at $BC$ contact $F=400\times 0.2=80N$

For slipping at $CG$ contact $F=600\times 0.1=60N$

So, as we increase the value of $F$, it first slip and at $F=60N$.

Thus $60N$ is the minimum force to start motion of any past of the system.

Hence (A) is the correct answer.