A machine of mass 20 kg is pushed across a horizontal floor by a force F directed at an angle of 37^ below the horizontal. The machine moves with a constant velocity. If the coefficient of kinetic friction between the machine and the floor is 0.5 , the magnitude of the pushing force F is (Given g = 10 m s ⁻² , 37^ = 0.6 , 37^ = 0.8 )

Step-by-step solution

Let N be the normal reaction from the floor. Resolving the pushing force F into horizontal and vertical components, we get F 37^ horizontally and F 37^ vertically downwards. For vertical equilibrium: N = Mg + F 37^ N = 20 10 + F(0.6) = 200 + 0.6F Since the machine moves with constant velocity, the net horizontal force is zero. Thus, the horizontal component of the pushing force equals the kinetic friction: F 37^ = _k N F(0.8) = 0.5(200 + 0.6F) 0.8F = 100 + 0.3F 0.5F = 100 F = 200 N Answer: 200 N