ME 211 Homework Set #2 Solutions Due: 1/23/2009 Problem 1 (3.37) Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has a radius of 5 ft, to fail at the support C. This requires a moment of M=80 lb ft to be developed at C. π«ππ = π«π βπ«π = 0 β 0 π’ + 5π ππ60Β°β 0 π£ + 5cos60Β°β 5 π€ = 4.3301π£β 2.5π€ ft π«ππ = π«π βπ«π = 6 β 0 π’ + 7 β 5π ππ60Β° π£ + 0 β 5cos60Β° π€ π
ππ = πΉπ΄π΅ π«ππ π« ππ = πΉπ΄π΅ 6 β 0 π’ + 7 β 5π ππ60Β° π£ + 0 β 5cos60Β° π€ 6 β 0 2 + 7 β 5π ππ60Β° 2 + 0 β 5cos60Β° 2 = πΉπ΄π΅(0.8538π’ + 0.3799π£β 0.3558π€) ππ = π«ππ Γ π
ππ ππ = πΉπ΄π΅ π’ π£ π€ 0 4.3301 β2.5 0.8538 0.3799 β0.3558 = πΉπ΄π΅(β0.5909π + 2.135πβ 3.697π) Magnitude of MC is therefore: ππΆ = πΉπ΄π΅ β0.59092 + 2.1352 + β3.6972 = 80 ο FAB=80/4.310=18.4518 lb Ans. Problem 2 (3.49) The hood of the automobile is supported by the strut AB, which exerts a force of F=24lb on the hood. Determine the moment of this force about the hinged axis y. (1) Position vectors: π«ππ = 4π π«ππ = β2π + 2π + 4π (2) Force vector: π
= 24 β2π + 2π + 4π (β2)2 + 22 + 42 = β9.8π + 9.8π + 19.6π ππ (3) Moment vector formulation: ππ = π π π 4 0 0 β9.8 9.8 19.6 = β78.4π + 39.2π ππ πππ¦ = ππ βπ = β78.4 ππ Ans. Problem 3 (3.61) A device called a rolamite is used in various ways to replace slipping motion with rolling motion. If the belt, which wraps between the rollers, is subjected to a tension of 15N, determine the reactive forces N of the top and bottom plates on the rollers so that the resultant couple acting on the rollers is equal to zero. Because of equilibrium: 15 75 = π(50πππ 30Β°) ο N=26.0 N Ans. RT RN B rOA x A y . . z O F Moy MO A Problem 4 (4.7) Determine the reactions at the pins A and B. The spring has an unstretched length of 80 mm. Solution: (1) Free Body Diagram: (2) Equation of Equilibrium: Problem 5 (4.16) If the wheelbarrow and its contents have a mass of 60 kg and center of mass at G, determine the magnitude of the resultant force which the man must exert on each of the two handles in order to hold the wheelbarrow in equilibrium. Solution: (1) Free Body Diagram: (2) Equations of Equilibrium: Problem 6 (4.18) The worker uses the hand truck to move material down the ramp. If the truck and its contents are held in the position shown and have a weight of 100lb with center of gravity at G, determine the resultant normal force of both wheels on the ground A and the magnitude of the force required at the grip B. Solution: (1) Draw the Free Body Diagram: (2) Write equilibrium equations and solve for unknowns: First, decompose N according to x and y direction: πΉπ = 0 ο πΉπ΅π + ππ ππ30Β°= 0(1) πΉπ = 0 ο πΉπ΅π + ππππ 30 βπΊ = 0 (2) Then, decompose N and G according to what FBD was showing , in this way, distances from the point to the force vector is easier to find. ππ΅ = 0 βπΊ1ππΊ1 βπΊ2ππΊ2 + π1ππ1 + π2ππ2 = 0 30ΒΊ βπΊπππ 30Β°β 1.5 + 1 βπΊπ ππ30Β°β 1.5 + 2 + ππ ππ30Β°β 0.5 + ππππ 30Β°β 1.5 + 2 + 1.75 = 0 (3) Where G= 100. Solve for N in Eq(3), N=81.6lb Ans From Eq (2), FBY=29.33 lb From Eq(1). FBX=-40.8 lb πΉπ΅ = πΉπ΅π2 + πΉπ΅π2 = 29.332 + 40.82 = 50.25 ππ Ans. Problem 7 (4.29) The rigid beam of negligible weight is supported horizontally by two springs and a pin. If the springs are uncompressed when the load is removed, determine the force in each spring when the load P is applied. Also, compute the vertical deflection of end C. Assume the spring stiffness k is large enough so that only small deflections occur. Solution: Problem 8 (4.31) The smooth pipe rests against the wall at the pints of contact A, B, and C. Determine the reactions at these points needed to support the vertical force of 45 lb. Neglect the pipeβs thickness in the calculation. Solution: (1) Draw the Free Body Diagram: (2) Write the equilibrium equations: Problem 9 (4.38) He shaft is supported by journal bearings at A and B. A key is inserted into the bearing at B in orfer to prevent the shaft from rotating about and translating along its axis. Determine the x, y, z components of reaction at the bearings when the 600-N force is applied to the arm. The bearings are in proper alignment and exert only force reactions on the shaft. Solution: (1) Draw the free body diagram: (2) Write equilibrium equations and solve for unknowns. Jinjin Ma