**A father and his son are supposed to carry a load of 500 N. They decided to use a uniform pole that is 4.0 m long and weighs 60 N. The father and his son support the pole at the ends. Where should the load be placed so that the father supports three times as much as his son?**

Posted by Daniel

Answer:

Let’s represent the situation be the folowing diagram

Let the force exerted by son be f

Then the force exerted by father = 3f

Therefore the total force exerted =Total weight = 560N = 4f

Which gives f = 140 N, the force exerted by son

3f=420 N, the force exerted by father

Applying the principle of moments about the point where father holds the rod,

420 x 0 – 500 x *x* – 60 x 2 + 140 x 4 = 0

==> **x**** = 440/500 = 0.88 m** from the end where father holds the load

(If there is any problem you find in the solution given, please post as comment to this post.)

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