Question

PROBLEM 8.23 The 10-lb uniform rod AB is held in the position shown by the force P. Knowing that the coefficient of static friction is 0.20 at A and B, determine the smallest value of P for which equilibrium is maintained. 15 i 7.5 in. 8 in. Fig. P8.23

Answer 1

Sketch the free body diagram of the rod as follows: 15 in. 7.5 in. (F), 8in. Apply the equilibrium equations. Calculate the forces along X-direction. F =0 (F) +P-Ng =0 UNA +P-N3 = 0 Here, u, is the coefficient of friction at point A, P is the load applied, N and N is the normal reaction force at point A and B. Substitute 0.2 for u. 0.2NA +P-N, = 0 Ng = 0.2N A +P Calculate the forces along y-direction. EF, =0 (F:), +NA-W =0 UzNz+N -W = 0 Here, W is the weight of the rod, u, is the coefficient of friction at point B. Substitute 0.2 for u, and 10 lb for W. 0.2N2 +N -10 =0) NA=10-0.2N3 Substitute the equation (2) in equation (1). N3 = 0.2NA+P Ng = 0.2(10 – 0.2N3) +P Ng = 2 – 0.04N3 +P 1.04N3 = 2 + P Ng =1.923 +0.96P Calculate the moment about point A. M.=0 (N3 *15)+((F;), 8)-(W x 4)-(Px7.5)=0 (N, *15)+(UxN, 8)–(W x 4)–(Px7.5)=0 Substitute 0.2 for y, and 10 lb for W. (Ng *15)+(0.2N3 *8)-(10x 4)-(Px7.5)= 0 16.6N2-40–7.5P=0 Substitute the equation (3) in equation (4). 16.6(1.923+0.96P)-40 - 7.5P=0 31.922 +15.936P - 40 -7.5P = 0 8.436P = 8.078 P=0.95716 ...... (4)