Question 1207522
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A uniform plank PQ of length 20m weighing 55N on a ceiling with an inextensible wire 
3m away from P. To keep the plank horizontally, it is made to rest on a support 8m from Q. 
Calculate the tension T on the wire and the weight W exerted on the support.
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<pre>
Let the tension be T newtons, and let W be the support force at the support point.

Then we have this equilibrium equation

    T + W = 55  newtons.    (1)


Second equation is equality of rotation moments around the center of the plank.


The leg of the force T is 20/2-3 = 7 meters; the leg of the force W is 20/2-8 = 2 meters.

The rotation moment of the force T is 7*T N*m.  The rotation moment of the force W is 2*W N*m.


The equation for rotation moments is

    7T = 2W.    (2)


So, we have two equations, (1) and (2), for two unknowns, T and W.


To solve, express T = 55-W from (1) and substitute it in equation (2).  You will get

    7*(55-W) = 2W,

    385 - 7W = 2W

    385 = 2W + 7W

    385 = 9W

    W = 385/9 = 42.778 Newtons  (rounded).


Hence, force T is  T = 55-42.778 = 12.222 Newtons.


<U>ANSWER</U>.  The tension is 12.222 N.  The support force is 42.778 N.
</pre>

Solved.


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Important post-solution notice


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Considering rotation moments around the central point of the plank, 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;we remove the influence and the contribution of the own weight of the plank

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;to rotation moments. It is where the condition is used that the plank is uniform.