Question

The figure shows a system of two blocks of $10kg$ and $5kg$ mass, connected by ideal strings and pulleys. Here the ground is smooth and the coefficient of friction between the two blocks is$\mu =0.5$. A horizontal force $F$ is applied on a lower block as shown. The minimum value of $F$ required to start sliding between the blocks is : (Take $g=10m/s^2)$

• A. $12.5N$
• B. $25N$
• C. $50N$
• D. $100N$

Answer 1

The correct option is B $25N$

In the condition that sliding between the blocks is just achieved, the friction will have its maximum value (i.e. $f=\mu N_1$)

For the $5kg$ block:

$f-T=5a_1$, along x-axis.

$N_1+T+T=50$, for equilibrium along y-axis.

Similarly, for the $10kg$ block:

$F-f-T=10a_2$, along x-axis.

Since this is the limiting case, the acceleration of the blocks will be zero:

$\therefore f-t=0\Rightarrow f=T$, and

$\therefore F_{min}-f-T=0\Rightarrow F_{min}=2T$

Since we know that $f=\mu N_1=0.5N_1$, we get $N_1=2f=2T$

Substituting this in $N_1+2T=50$, we get $T=12.5N$

And thus, $F_{min}=2T=25N$.