Question #09151

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Question #09151

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Answer 1 · 2016-12-11T17:46:11

$M = 34.9677$ $M=34.9677$ [kg]

Explanation:

Calling $T$ $T$ the tension on the rope, $N_{1}$ $N_1$ the normal force at the man floor, $N_{2}$ $N_2$ the normal force at the mass location, $μ_{s}$ $mu_s$ the friction coefficient, $m$ $m$ the man mass $θ$ $theta$ the angle at the rope and $g$ $g$ the gravity, we have

Horizontally

$T cos (θ) - μ_{s} N_{1} = 0$ $T cos(theta)-mu_s N_1=0$

Vertically

$N_{1} - m g + T sin (θ) = 0$ $N_1-m g + T sin(theta)=0$ and $T + N_{2} - M g = 0$ $T+N_2-M g=0$

but at the problem conditions we have $N_{2} = 0$ $N_2=0$ so we have

${(T cos(theta)-mu_s N_1=0),(N_1-m g + T sin(theta)=0),(T-M g=0):}$

Solving for $T,N_1,M$ we obtain

$((T = (g m mu_s)/(Cos(theta) + mu_s Sin(theta))),(N_1 = (g m Cos(theta))/(Cos(theta) + mu_s Sin(theta))),(M = (m mu_s)/(Cos(theta) + mu_s Sin(theta))))$

So finally

$M = (m mu_s)/(Cos(theta) + mu_s Sin(theta))=34.9677$

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Question #09151

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