Question

If the uniform concrete block has a mass of 500 kg , determine the smallest horizontal force P needed to move the wedge to the left . The coefficient of static friction between the wedge and the concrete and the wedge and the floor is , _s= 0.3 . The coefficient of static friction between the concrete and floor is '_s= 0.5 .

                   

{\displaystyle \mu \,}

{\displaystyle \mu \,}

Answer 1

Solution

Free - Body Diagram . Since the wedge is required to be on the verge of sliding to the left , the frictional forces FB and FC on the wedge must act to the right such that F_B =  _sN_B = 0.3N_B and F_C = _sN_C= 0.3N_C.

Equations of Equilibrium . Referring to the free - body diagram of the concrete block shown in Fig . a 

Since F_A < ( F_A ) max = μ'_s N_A = 0.5 ( 2506.40 ) = 1253.20 N , the concrete block will not slip at A. Using the result of N_B and referring to the free - body diagram of the wedge shown in Fig . b