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The 550 N box (fig.) rests on a horizontal plane for which the coefficient of kinetic friction $\mu_x=0.32$ .

If the box is subjected to a 400 N towing force as shown, find the velocity of the box in 4 seconds starting from the rest.

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Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 04

Years : MAY 2015

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$∑Fy = 0$ (because the box will not move vertically at all)

$N + 400 \sin30 – 550 = 0 \\ N= 350N $

$ ∑Fx = m \space a_x$ (by D’Alembert’s principle)

$400 \cos30 – 0.32 \times 350 = 550/9.8 \times a_x\hspace {2cm} (W = mg) \\ A_x = 4.18 m/s^2$

Using, $v= u + at \\ V= 0 + 4.18 \times 4 = 16.7 m/s$

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