Problem: A block weighing 1500 kg is to be raised by means of a 10 degree wedge. Assuming the coefficient of friction between all the contacting surfaces to be 0.3, determine what minimum horizontal force P, should be applied to raise the block? Appeared in Bihar University in the year 1974. Solution: Given, Weight of the block W = 1500 kg, Angle of the wedge = alpha = 10 degree, Friction coefficient, mu = 0.3 = tan phi, phi = 16 degree 42` Some thinking shows that when the block tends to move upwards, sliding is initiated across the surfaces in contact. We can try solving theproblem graphically first. First of all
let’s consider the block to be in equilibrium. We take any point l and execute a vertical line lm = 1500 kg to a certain scale. Through m we draw a line parallel to the reaction R2 and through l we draw a line parallel to the reaction R3. We assume both these lines to meet at n. Similarly, we assume the wedge to be at the state of equilibrium and draw a line through m parallel to the horizontal force P and draw a line through n parallel to the reaction R1. We assume these two lines to be meeting at O. Now measuring mo, we get the applied force as 1417.5 kg.