Frictional Forces on Flat Belts Learning Goal: To be able to analyze the equilibrium of systems subjected to frictional forces at belts and surfaces. When engineers design belt drives and band brakes, they engineer them to withstand the frictional forces that develop between the belts or bands and their contacting surfaces. These frictional forces lead to tensions on each side of the contacting surfaces that are unequal: the segment aligning with the motion of the belt or band has a greater tension. Part A A cable is attached to a 1.20kg block A, is looped over a fixed peg at C, and is attached to plate B. (Figure 1) The coefficient of static friction between the plate and the block is muA = 0.260, the coefficient of static friction between the plate and the inclined plane is muB = 0.450, and the coefficient of static friction between the cable and the peg is muc = 0.360. If the plane's angle is 30.0 degrees. what is the maximum mass that plate B can have such that it does not slide down the plane? Express your answer numerically in kilograms to three significant figures. mB = kg Part B Blocks A and B have a masses of 100kg and 270kg , respectively.(Figure 2) The coefficient of static friction between A and B and between B and C is 0.270. The coefficient of static friction between the rope and peg E is 0.560. Pulley D rotates freely, and P = 29.0N . If B = 60.0degrees what is the smallest magnitude, T, of tension, T, that causes block B to move? Express your answer numerically in kilonewtons to three significant figures. T = N Part C Block A has a mass of mA = 46.0kg and rests on a flat surface. (Figure 3) The coefficient of static friction between the block and the surface is muA = 0.290. The coefficient of static friction between the rope and the fixed peg B is 0.360. The width of the block is d = 0.270m. Find the greatest mass. mc. that weight C can have such that block A does not move. Express your answer numerically in kilograms to three significant figures. mC = kg