In the arrangement shown in figure, the inclined plane is 130 cm long and its upper end is 50cm above the level of the lower end. The block m 2 rests on the plane and has a mass of 60 kg . The block

Question

In the arrangement shown in figure, the inclined plane is 130 cm long and its upper end is 50cm above the level of the lower end. The block m 2 ​ rests on the plane and has a mass of 60 kg . The block m 1 ​ has a mass 200 kg. The coefficient of static friction between the two blocks is 0.50; the coefficient of sliding friction between the lower block and the plane is 0.33 . A force F upward and parallel to the plane is applied to the lower block. What is the maximum value of F before the two blocks move with respect to each other?

TestHub · Accepted Answer

Upper block's maximum acceleration (friction between blocks): $a_{1 (ma x)} = g (μ cos θ - sin θ) = \frac{10}{13}$ m/s $^{2}$ , with $sin θ = \frac{5}{13}$ , $cos θ = \frac{12}{13}$ .

For both blocks together: $F - (m_{1} + m_{2}) g (sin θ + μ_{0} cos θ) = (m_{1} + m_{2}) a_{1 (ma x)}$ .

$F = 260 \times 10 \times \frac{5 + 0.33 \times 12}{13} + 260 \times \frac{10}{13} \approx 2000$ N - (A)

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