Question

In the figure given, the system is in equilibrium. What is the maximum value that $W$ can have if the friction force on the $40N$ block cannot exceed $12.0N$

• A. $3.45N$
• B. $6.92N$
• C. $10.35N$
• D. $12.32N$

Answer 1

The correct option is B $6.92N$

Maximum value of friction force ($F_{max}$) is given as $12.0N$. Thus, for equilibrium of the block along the horizontal axis, we have:

$T-F_{max}=0$

$\Rightarrow T=F_{max}=12.0N$

From the 2nd FBD, it is clear that $T_1$ is the same force as $T$. Thus, for equilibrium of the junction point:

$T_2\cos {30}^{\circ }-T_1=0\Rightarrow T_2=\frac{T_1}{\cos {30}^{\circ }}=\frac{T}{\cos {30}^{\circ }}$

And for the vertical axis: $W=T_2\sin {30}^{\circ }$

$\Rightarrow W=T\tan {30}^{\circ }=\frac{12}{\sqrt{3}}\approx 6.92N$