1) A force P of magnitude 90 lb is applied to a member ACE, which is supported b

Question

1) A force P of magnitude 90 lb is applied to a member ACE, which is supported by a frictionless pin at Dand by a cable ABE. Sincethe cable passes over a small pulley at B, the tension may be assumed to be the same pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. Forthe case when a= 3in., determine (a) the tension in the cable, (b) the tension in the cable, (b) the reaction at D.

                    Correct answers are
                

                    1) (a)195.0 lb (b) 255 lb C= 45.0 degrees

Help me get to these answers

A force P of magnitude 90 lb is applied to a member ACE, which is supported by a frictionless pin at D and by a cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a= 3in., determine (a) the tension in the cable, (b) the tension in the cable, (b) the reaction at D.Correct answers are (a)195.0 lb (b) 255 lb C= 45.0 degrees

Explanation / Answer

vertical componet of tension at A = (12/13) * T

horizontal componentof tension at A = ( 5/13) * T


taking moment of all forces abount D

T*3 + 90*9 = (5/13)*T*9 + (12/13)*T*7

so T = 117 lb


so tension in the cable = 117 lb


b)
Let the reaction at D be Dx and Dy

for equilibrium of all forces in the x-direction

Dx - T - (5/13)*T + P = 0

so Dx = 72 lb


for equillibrium in y direction

Dy + (12/13)*T = 0

so Dy = - 108 lb

so reaction at D = ( 72 i^ - 108 j^ )

magnitude of net reaction at D = sqrt ( 108^2 + 72^2 ) = 129.7998 lb

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