Question

Determine the horizontal force P required to start moving the 500 N block as shown in figure 3 up the inclined surface. Assume μs= 0.3.

Figure 3.

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Answer 1

$P -wsin θ -f =0 \\ P = wsin θ +f \\ P = wsin θ - \mu_k w cos θ \\ P = 500 \times sin(30°) + 0.3 \times 500 \times cos(30°) \\ P = 250 + 129.9 \\ P = 379.9 \;N$