Question

# In previous problem, what is the smallest force $F$ necessary to hold the box stationary on the inclined surface ?

Solution

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At impending slip, the sum of the forces parallel to the surface is:

\textit$\color{#c34632} \sum F_{P}=f-W_{P}=0$

From which:

\textit$\color{#4257b2} f=W_{P}=\mu_{s}(F+W \cos \alpha)$

And,

\textit$\color{#4257b2} W_{P}=w \sin \alpha$

Equate and solve:

\textit$\color{#4257b2} F=W \big( \frac{\sin \alpha}{\mu_{s}}- \cos \alpha \big)=30 \big( \frac{ \sin 20 }{\mu_{s}}- \cos 20 \big) = \boxed{23.1 \ N}$