Question

If coefficient of friction between the surface of 25 kg block and fixed vertical wall is 0.4 , and that between both the blocks is also 0.4. All the shown pulleys are light and frictionless; strings are light, perfectly inextensible and smooth. Then, (Take $g=10m/s^2)$

• A. to have the complete system in static equilibrium, the magnitude of minimum horizontal force F towards right (as shown) is equal to $\frac{125}{4}$N.
• B. to have the complete system in static equilibrium, the magnitude of minimum horizontal force F towards right (as shown) is equal to 125 N.
• C. in the equilibrium state, the direction of frictional force exerted by vertical wall on 25 kg block and the direction of frictional force exerted by 25 kg block on 15 kg block, downwards & upwards, respectively.
• D. in the equilibrium state, the direction of frictional force exerted by vertical wall on 25 kg block and the direction of frictional force exerted by 25 kg block on 15 kg block, upwards & downwards, respectively.

Answer 1

Answer: (A), (C)

The correct options are
A to have the complete system in static equilibrium, the magnitude of minimum horizontal force F towards right (as shown) is equal to $\frac{125}{4}$N.
C in the equilibrium state, the direction of frictional force exerted by vertical wall on 25 kg block and the direction of frictional force exerted by 25 kg block on 15 kg block, downwards & upwards, respectively.

FBD of the system is shown in below figure.
For horizontal equilibrium of the blocks, $N_1=N_2=F....\left(1\right)$
$f_1=\mu N_1,f_2=\mu N_2$
Therefore, from (1), $f_1=f_2=\mu F....\left(2\right)$
For $15kg$ block, $T_1+f_1=15g.....\left(3\right)$
For $25kg$ block, $2T_1=f_2+25g+f_1......\left(4\right)$
Using (2) and (3), (4) becomes, $30g-2\mu F=\mu F+25g+\mu F$
$4\mu F=5g,F=\frac{50}{4\left(0.4\right)}=\frac{125}{4}N$
Now as the value of force $F$ is increased to $25N$, the magnitude of frictional forces do not change.
Therefore, (A) and (C) are the correct choices.