# Question #09151

Dec 11, 2016

$M = 34.9677$ [kg]

#### Explanation:

Calling $T$ the tension on the rope, ${N}_{1}$ the normal force at the man floor, ${N}_{2}$ the normal force at the mass location, ${\mu}_{s}$ the friction coefficient, $m$ the man mass $\theta$ the angle at the rope and $g$ the gravity, we have

Horizontally

$T \cos \left(\theta\right) - {\mu}_{s} {N}_{1} = 0$

Vertically

${N}_{1} - m g + T \sin \left(\theta\right) = 0$ and $T + {N}_{2} - M g = 0$

but at the problem conditions we have ${N}_{2} = 0$ so we have

$\left\{\begin{matrix}T \cos \left(\theta\right) - {\mu}_{s} {N}_{1} = 0 \\ {N}_{1} - m g + T \sin \left(\theta\right) = 0 \\ T - M g = 0\end{matrix}\right.$

Solving for $T , {N}_{1} , M$ we obtain

$\left(\begin{matrix}T = \frac{g m {\mu}_{s}}{C o s \left(\theta\right) + {\mu}_{s} S \in \left(\theta\right)} \\ {N}_{1} = \frac{g m C o s \left(\theta\right)}{C o s \left(\theta\right) + {\mu}_{s} S \in \left(\theta\right)} \\ M = \frac{m {\mu}_{s}}{C o s \left(\theta\right) + {\mu}_{s} S \in \left(\theta\right)}\end{matrix}\right)$

So finally

$M = \frac{m {\mu}_{s}}{C o s \left(\theta\right) + {\mu}_{s} S \in \left(\theta\right)} = 34.9677$