Question

If coefficient of friction between all surfaces is 0.4, then find the minimum force F to have equilibrium of the system. (Take g = $10\frac{m}{s^2}$ )

Answer 1

Tension acting on the string as shown in figure indicates $25kg$ load will be lifted up by this pulley system. Hence $25kg$ load will move upwards, while $15kg$ load will move downwards.

Free body diagram of both the blocks showing the action of different forces are given in the figure.

Friction force against wall for $25kg$ block $=f_W$

Friction force against the contacting surfaces between block that act on $25kg$ block $=f_B$

Reaction force of $f_B$ that acts on $15kg$ block $=f_{B1}$

Friction force against the contacting surfaces between block that act on $15kg$ block $=f_A$

Reaction force on $f_A$ that acts on $25kg$ block $=f_{A1}$

We have,

$f_W=f_A=f_{A1}=f_B=f_{B1}=\mu F$

where $\mu$ is friction coefficient and $F$ is applied force.

For equilibrium of $15kg$ block $:T=2\mu F=15g$

For equilibrium of $25kg$ block $:2T=25g+3\mu F$

Solving the above equations, we get,

$F=\frac{25}{14}g=17.5N$