Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08

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Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08

8-1.The mine car and its contents have a total mass 0Í 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and

Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08 d the tracks is = 0.4 when the wheels arc locked, find the normal force acting on the front wheels at /Ỉ and the rear wheels at /1 when the brakes at

both A and B are locked. Does the car move?SOLUTIONEquations of Equilibrium: The normal reactions acting on the wheels at (zt and are independent as t Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08

o whether the wheels are locked or not. Hence, the normal reactions acting on the wheels are the same for both cases.C+XM, = 0;NA (1.5) + 10(1.05) - 5

Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08

8.86(0.6) = t)NA = 16.544 kN = 16.5 kNAns.0Nb + 16.544 - 58.86 = 0Nb = 42.316 kN = 42.3 kNAns. ,When both wheels at A an

8-1.The mine car and its contents have a total mass 0Í 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and

Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08 els do not slip. Thus, the mine car docsnot move.'c-AbsP Jy_________ ______________________khothlwPn.comV <0 . 'Tr -A©2013 Pearson Education Inc.. upp

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Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08

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8-1.The mine car and its contents have a total mass 0Í 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and

Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08 tion and each is tightened so that it is subjected to a tension of 4 kN. Ute coefficient of static friction between the plates is p, = 0.4.SOLUTIONFre

e-Body Diagram: The normal reaction acting on the contacting surface is equal to the sum total tension of the bolts. Titus, iV = 4(4) kN = 16 kN. When Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08

the plate is on the verge of slipping, the magnitude of the friction force acting on each contact surface can be computed using the friction formula

Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08

F = HịN = 0.4(16) kN. As indicated on the frcc-body diagram of the upper plate. F acts to the right since the plate has a tendency to move to the left

8-1.The mine car and its contents have a total mass 0Í 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and