For a block of weight W on a horizontal surface with mu_s=0.35, the minimum horizontal force to initiate sliding is:

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Mechanical Engineering QUESTION #2542

Question 1

For a block of weight \(W\) on a horizontal surface with \(\mu_s=0.35\), the minimum horizontal force to initiate sliding is:

\(P_{\min}=0.35\,W\)✔️
\(P_{\min}=W/0.35\)
\(P_{\min}=0.35\,W^2\)
Cannot be determined without contact area

Correct Answer Explanation

At impending slip, friction reaches its limit: \(F_{\max}=\mu_s N=\mu_s W=0.35W\). Hence the applied force must equal \(P=0.35W\). Note that friction force is independent of contact area (Amontons' law).

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