Question

1. An 8.25 kg sits at rest on an inclined plane with an angle of 18.8 degrees, 11.2 cm from the base of the inclined plane. The coefficient of static friction between the block and inclined plane is 0.55 The incline plane is slowly adjusted until the block starts sliding down the plane. At what angle will the block start moving?

2. A 12.0 kg sign hangs from the end of a 2.00 m long horizontal beam. What is the tension in the cable? The beam has a mass of 5.00 kg

Answer 1

1) If it starts to side then the static friction is balanced by gravitational force mgsin 0 = UN = umg cose jul = tane O= tan-'(u) = tan- (0.55) = 28.8 2) If the system is in equilibrium, the torque is zero T sin e-mg s -m,gd = 0 L T sin = mg + + ,gd mig T=- ( +magd sin e