You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp (?Fig. E5.33?). Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800. (a) What force do you need to exert to accomplish this? (b) What are the magnitude and direction of the friction force on the upper box?

Solution 31E Step 1: In this question, we to find find The force needed to pull the boxes down the ramp The magnitude and direction of the friction force for the above box Let us consider the data given Mass of the box 1 = m = 32 kg 1 Mass of box 2 m = 45 kg 2 Width of the ramp w = 4.75 m Length of the ramp l = 2.50 m Velocity of the box down the ramp v = 15 cm /s Coefficient of kinetic friction between ramp and lower box = 0.444 f Coefficient of static friction between two boxes = 0s800 Step 2 : Let us find the angle of the slope It is given using 1 = tan (l/w) Substituting the values we get = tan (2.5/4.75) = tan (0.5263) 0 = 27.75 0 Hence we have obtained the angle of the ramp as = 27.75 Step 3 : Let us find the sliding force down the ramp This is obtained using F = m × sin f We have the masses as m 1 m = m2 32 kg 45 kg = 77 kg Substituting the values we get 0 2 F f 77 × sin × 27.75 × 9.8 m/s F f 77 × 0.4656 × 9.8 m/s 2 2 F f 35.85 N × 9.8 m/s F f 351.35 N Hence we have frictional force as 351.35 N