Question

In the diagram shown, the blocks A, B and C weigh 3 kg, 4 kg and 8kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a mass less rigid rod fixed to the wall while B and C are connected by a flexible cord passing round a fixed frictionless pulley. Assuming that the arrangement shown in the figure (i.e., B on C and A on B) is maintained all throughout, the force necessary to drag C along the horizontal, to the left at constant speed is (in N)

• A. 8
• B. 7
• C. 5
• D. 9

Answer 1

Answer: (A)

8

When C moves to left , the block B moves to right.
Tension in string$=$ sum of frictional forces at upper and lower surfaces of B
$\Rightarrow T=\mu \left(W_A\right)+\mu \left(W_{A+B}\right)$
$\Rightarrow T=0.25\times 3+0.25\times (3+4)=2.5N$

When C slides to left, required force F$=$ sum of tension T and frictional forces at upper and lower surfaces of C
$=T+0.25\left(W_{A+B}\right)+0.25\left(W_{A+B+C}\right)$
$=2.5+0.25\left(7\right)+0.25\left(15\right)$
$=8N$