An applied force F acts on three boxes as shown in the figure. The coefficient o
ID: 1707255 • Letter: A
Question
An applied force F acts on three boxes as shown in the figure. The coefficient of kinetic friction mu = -0.10 and is the for each box The boxes are initially at rest. The applied force IS increased in flora mo TV boxes start to move when the applied force just exceeds a 120N.Sbowing all yow work. find the coefficient of static between the lower surface of the boxes the surface on which they rest (assumed lo be the same for each box). Once the boxes are moving force, the applied force is adjusted so that the boxes air pushed at a constant speed to the right of 2m/s. Showing all your work. calculate (be magnitude of the force exerted by box #1 on box #2. The applied force is adjusted so that the boxes accelerate to the right at a rate of 1.0m/s. Showing all your work calculate the magnitude of the force FExplanation / Answer
The given masses are m1 = 15kg, m2 = 40kg and m3 = 25kg (15) The applied force F = 120N then the coefficient of static friction = F/mg = 120 / (15 + 40 + 25 ) (9.8) = 0.15 (16) When they are moving with constant speed, the acceleration is zero so Fnet = F - f1 -F21 = 0 F = f1 + F21 for second block F12-F32 - f2 = 0 and for third block F32 - f3 = 0 F32 = m3g = (0.15)(25kg)(9.8) = 36.75 N then F12 = 36.75 + ( m2g) = 95.55 N Therefore the force exert by first box on second box is 95.55N (17) When they are moving acceleration a = 1.0m/s^2 so Fnet = F - f1 -F21 = 0 m1 a = F - f1 + F21 15N = F - f1 + F21 for second block F12-F32 - f2 = 40 N and for third block F32 - f3 = 25N F32 = m3g + 25 = (0.15)(25kg)(9.8) +25 = 61.75 N then F12 = 61.75 + ( m2g) + 40 = 160.55 N Therefore the applied force F = 15 + (0.15)(15)(9.8) + 160.55 = 197.6 N
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