A mass of 3.10 kg is suspended from the end of a thin, uniform, horizontal rod w

Question

A mass of 3.10 kg is suspended from the end of a thin, uniform, horizontal rod with a mass of 2.70 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.

The figure is at this link http://imageshack.com/a/img673/628/p4R7EC.png

1. Calculate the tension in the cable.

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.

Explanation / Answer

Here ,

let the string makes an angle theta with the rod ,

theta = arctan(5/12)

theta = 22.6 degree

Now, let the tension in the string is T ,

Balancing the moment about point of contact on rod ,

3.10 * 9.8 * 12 + 2.70 * 9.8 * 6 - T * sin(22.6) * 12 = 0

solving for T

T= 113.5 N

the tension in the string is 113.5 N

2) now, let the hozizontal force acting is Fx

Fx - T * cos(22.6) = 0

Fx = 113.5 * cos(22.6) = 0

Fx = 104.8 N

Now , in the vertical direction ,

Fy - (3.10 + 2.70)* 9.8 + 113.5 * sin(22.6) = 0

Fy = 13.22 N

Now,

minimum value of the coefficient of static friction = Fy/Fx

minimum value of the coefficient of static friction = 13.22/104.8

minimum value of the coefficient of static friction = 0.126

the minimum value of the coefficient of static friction is 0.126

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