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A rigid, vertical rod with a mass of 3.0 kg simply rests on the floor and is held in place by static friction. The coefficient of static friction between the rod and the floor is1/7. The rod also has a wire connected between its top end and the floor, as shown in Figure 11-31. A horizontal force F is applied at the midpoint of the rod.
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Added Sun, 23 Aug '15
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Forces on the rod:

reaction N upwards at bottom end

force F at middle

Tension T in the string at 45 degree from the vertical

frictional force Fr on the bottom end and the direction is to the left side.

weight mg downwards

vertical force balance

mg -N + Tcos 45 = 0

so N = mg +Tcos 45 .............(1)

horizontal force balance

Fr -F +Tsin45 = 0

uN -F +Tsin 45 =0

substituting the value of N from (1)

(1/7)[mg+Tcos45] -F +Tsin45 =0

(1/7)[3*9.8 +T*0.7] -F +T*0.7 =0

4.2 +0.8T =F

T= (F-4.2)/0.8

moment balance about top end

F*(L/2) - Fr*L =0

F/2 = Fr

F/2 = F- Tsin45

F/2 = Tsin45

F= 2Tsin45 = 2*0.7*(F- 4.2)/0.8

F = 1.75F - 7.35

0.75 F = 7.35

F = 9.8

so maximum F is 9.8 N

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Added Sun, 23 Aug '15
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