This force can either push the block upward at a constant velocity or allow it t

Question

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The magnitude of the force is different in the two cases, while the directional angle is the same. Kinetic friction exists between the block and the wall, and the coefficient of kinetic friction is 0.280. The weight of the block is 57.0 N, and the directional angle for the force Upper F Overscript right-arrow EndScripts is = 48.0°. Determine the magnitude of Upper F Overscript right-arrow EndScripts when the block slides (a) up the wall and (b) down the wall.

Explanation / Answer

48° = angle of applied force wrst vertical

Fup = applied force UP the wall at angle 48°

Fdn = applied force DOWN the wall at angle 48°

FF = force of kinetic fricition = 0.28 x NF (direction opposite to movement)

NF = normal force = applied force x sin 48

W = weight of block = 57 N {downward}

Fnet = net force applied to block parallel to wall = 0 {ref constant speed motion}

a) UP

Fnet = (Fup)(cos 48) - FF - W

0 = (Fup)(cos 48) - (0.28)(Fup)(sin 48) - 57

57 = Fup [cos 48 - 0.28(sin 48)] = 0.461(Fup)

Fup = 57/0.461 = 123.63 N ANS (a)

b) DOWN

Fnet = (Fdn)(cos 48) + W - FF

FF = (Fdn)(cos 48) + W = 0.669(Fdn) + 57

(0.28)(Fdn)(sin 48)] = 0.669(Fdn) + 57

0.208(Fdn) = 0.669(Fdn) + 57

-0.4609(Fdn) = 57

Fdn = - 57/0.4609 = -123.63 N {neg means no real ans - see comment}

Comment: answer means APPLIED FORCE = Fdn must be directed 180° to that stated. Thus it is impossible to have an applied force at 48° be directed down wall and meet the other conditions of the described motion. ANS

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