1. Problem 8.150 Block A of mass 12 kg and block B of mass 6 kg are connected by a cable that passes over pulley C which can rotate freely. Knowing that the coefficient of static friction at all surfaces of contact is 0.12, determine the smallest value of P for which equilibrium is maintained. A B P C

2. Solving Problems on Your Own When the motion is impending and  s is known; you must find some unknown quantities, such as a distance, an angle, the magnitude of a force, or the direction of a force. Block A of mass 12 kg and block B of mass 6 kg are connected by a cable that passes over pulley C which can rotate freely. Knowing that the coefficient of static friction at all surfaces of contact is 0.12, determine the smallest value of P for which equilibrium is maintained. A B P C a. Assume a possible motion of the body and, on the free-body diagram, draw the friction force in a direction opposite to that of the assumed motion. Problem 8.150

3. Solving Problems on Your Own When the motion is impending and  s is known; you must find some unknown quantities, such as a distance, an angle, the magnitude of a force, or the direction of a force. Block A of mass 12 kg and block B of mass 6 kg are connected by a cable that passes over pulley C which can rotate freely. Knowing that the coefficient of static friction at all surfaces of contact is 0.12, determine the smallest value of P for which equilibrium is maintained. A B P C b. Since motion is impending, F = F m =  s N. Substituting for  s its known value, you can express F in terms of N on the free- body diagram, thus eliminating one unknown. c. Write and solve the equilibrium equations for the unknown you seek. Problem 8.150

4. Problem 8.150 Solution A P Assume a possible motion of the body and, on the free-body diagram, draw the friction force in a direction opposite to that of the assumed motion. TATA Impending Motion: Block A Block B N = P W A = m A g FAFA FAFA N = P B TBTB W B = m B g FBFB N = P Free-Body Diagrams

5. A P TATA Impending Motion: Block A Block B N = P W A = m A g N = P B TBTB W B = m B g F B =  s N N = P Free-Body Diagrams Since motion is impending, F = F m =  s N. Substituting for  s its known value, you can express F in terms of N on the free-body diagram, thus eliminating one unknown. F A =  s N F A = m s N Assume a possible motion of the body and,on the free-body diagram, draw the friction force in a direction opposite to that of the assumed motion. Problem 8.150 Solution

6.  F y = 0 : T A + 2F A - W A = 0  F y = 0 : T B - F B - W B = 0 A P TATA N = P W A = m A g N = P B TBTB W B = m B g F B =  s N N = P F A =  s N Write and solve the equilibrium equations for the unknown. + + T A + 2  s N - m A g = 0 T B -  s N - m B g = 0 T A = m A g - 2  s N T B = m B g +  s N But, T A = T B : m A g - 2  s N = m B g +  s N Problem 8.150 Solution

7. A P TATA N = P W A = m A g N = P B TBTB W B = m B g F B =  s N N = P F A =  s N Write and solve the equilibrium equations for the unknown. m A g - 2  s N = m B g +  s N (m A - m B ) g = 3  s N (12 kg - 6 kg) g = 3(0.12)N 6 g 0.36 N = = 16.667 g = 16.667(9.81 m/s 2 ) = 163.5 N Since P = N, we have P = 163.5 N Problem 8.150 Solution