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Horizontal lifting boom

A mass of 3.10 kg is suspended from the end of a thin, uniform, horizontal rod with a mass of 1.90 kg. As shown below, one end of the rod is in contact with a wall and is supported by a thin wire attached to the wall. Friction between the wall and rod keeps the rod from slipping.

1. Calculate the tension in the cable.
Note: the grid spacing in the figure is 10 cm, in both horizontal and vertical directions.

2.Calculate the minimum value of the coefficient of static friction between the wall and the rod which is required to keep the rod from slipping.
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Added Fri, 03 Jul '15
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a)
sum torque about end
mblock g*0.08 + m rod*g*0.04 - T*0.08*sin(theta)
sin(theta) = opp/hyp = 7/sqrt(7^2+7^2)
3.1*9.81*0.08 + 1.9*9.81*0.04 - T*0.08*7/sqrt(7^2+7^2)=0
T=56.188 N (answer)

b)
sum forces in the x
Fxwall - T cos theta = 
Fxwall = 56.188*8/sqrt(7^2+7^2)= 39.72
sum forces in the y
Fywall -3.1*9.81 - 1.9*9.81+79.89*7/sqrt(7^2+8^2)=0
Fywall = 9.327

u = Fywall/Fxwall = 9.322/39.72=0.2347
u = 0.2347 ( answer)
Edit
Added Fri, 03 Jul '15
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