A 2.00 kg block is held in equilibrium on an incline of angle = 50 degree by a h

Question

A 2.00 kg block is held in equilibrium on an incline of angle = 50 degree by a horizontal F applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is = 0.300, determine the following, the minimum value of F 23.16 Your response differs from the correct answer by more than 10%, Double check your calculations, N the normal force exerted by the incline on the block 30,347 Your response differs from the correct answer by more than 10%. Double check your calculations. N

Explanation / Answer

a)
Fcos 50 = mgsin50 - u(mgcos50 + Fsin50)
.6427876 F = 2x9.81x.766 - .3(2x9.81x.6427876 + F.766)

F = 12.8655 N



b)

normal force = mgcos50 + Fsin50
= 2x9.81xcos 50 + 12.8655 sin50

N =22.467 N

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