# How weight of an object divides into components

1. Aug 2, 2015

### fireflies

Let us suppose a rod is inclined on the ground such that its middle point is supported by a peg. It is a uniform rod and the point where it touches the ground is A, mid-point (also the centre of gravity) is C. The rod is inclined at angle theta with
horizon. So, we can divide the components of weight into
mgcos(theta) against the normal of the rod, andg sin (theta) along
the rod. Then how do we measure it if the peg was at any
other point D (I meant measure
the weight at D), which is not the centre of gravity? What are the conditions to support the rod then, when the angle is same? How frictional force is at the point touching the ground?

2. Aug 2, 2015

### Dr. Courtney

A diagram helps a lot for these problems.

3. Aug 2, 2015

### fireflies

Yes, I tried with a diagram.

The actual confusion started when I tried to solve this problem www.physicsforums.com/threads/newtons-third-law-confusion.825741/

There I got at C the force along the rod is mg cos theta which passes through the point touching ground. At that point I further divided the mg sin theta into components to find the frictional force and reaction force. Then, the question came, why not use weight here. I know the answer, it is because the weight acts along the centre of gravity. Then the question comes, what should I do then if the peg was some point above C, suppose at D? What would it change in frictional force then?

Well, I think the main problem I am having is with centre of gravity. How it works when we consider the other points of the object other than centre of gravity? Maybe the answers of the previous questions may give a clear view to the later ones