A box with mass m is dragged across a level floor with coefficient of kinetic friction µk by a rope that is pulled upward at an angle ? above the horizontal with a force of magnitude F. (a) In terms of m, µk, ?, and g , obtain an expression for the magnitude of the force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90-kg patient across a floor at constant speed by pulling on him at an angle of 25o above the horizontal. By dragging weights wrapped in an old pair of pants down the hall with a spring balance, you find that µk = 0.35. Use the result of part (a) to answer the instructor’s question.
Solution 38E Problem (a) To obtain the expression for Force to move the box with constant speed Step 1: The coefficient of kinetic friction = k Angle between the rope and floor = At a constant velocity the box in equilibrium. Since the box is in motion the floor must exert a kinetic friction force. The force exerted by the rope has vertical component and reduce n and f k Step 2: From the equilibrium condition f k = kn Step 3: From horizontal component of force F x = Fcos .n k 0 Fcos .nk= 0 Fcos = .n k--(1) Step 4: From horizontal vertical of force F = Fsin w + n = 0 y Fsin = w n ----(2) Step 5: Solving (1) and (2) We will get k.w F = cos + sinor k km.g F = cos + k sin Problem (b) Step 1 m = 90 kg = 25° = 0.35 k