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

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

written 7.8 years ago by

teamques10 ★ 64k

• modified 3.9 years ago

. 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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engineering mechanics

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written 7.8 years ago by

teamques10 ★ 64k

enter image description here

$∑Fy = 0$ (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)

$$400 \cos30 – 0.32 \times 350 = 550/9.8 \times a_x (W = mg)$$

$A_x = 4.18 m/s^2 \\ Using, v= u + at \\ V= 0 + 4.18 \times 4 = 16.7 m/s$

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