Here, two unlike unequal parallel forces, P and Q are acting on the body at point A and B respectively. Now, If I find the resultant of these two forces, the resultant force, R acts at point C which is away from the body. So, if the resultant acts on a point outside of the body, how does it have any impact on the body? What's the significance of point C?
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$\begingroup$ I think you are somewhat confused. When several forces are applied on a rigid body, it experiences essentially a net force and a net torque. The net force is equivalent to a single force acting on the com of body. The net torque can be calculated either about com or icr. If you're using com, the total motion is superposition of translational and rotational motion. If ure using icr you just have rotational about icr and don't need the net force any more. $\endgroup$– Abhijeet MelkaniCommented May 8, 2017 at 8:15
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$\begingroup$ So, doesn't point C have any significance here? $\endgroup$– Kawsar AhmedCommented May 8, 2017 at 8:21
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$\begingroup$ No. It doesn't. $\endgroup$– Abhijeet MelkaniCommented May 8, 2017 at 8:28
1 Answer
Point $C$ has no significance as far as the diagram of the object is concerned.
The diagram that you have drawn shows the points of application and directions of two forces on an abject.
The length of any arrow that you draw to represent forces relates to the magnitude of the forces in newtons not metres.
When drawing the force diagram you could use any scale where a force of a certain magnitude is represented by an arrow of of certain length, eg 1 cm is
equivalent to 1 newtons or 10 newtons, or 100 newtons etc.
So you are superimposing a force diagram over a diagram of an object which has two forces acting on it.
The net force acting on the object is $\vec P- \vec Q$ and there will also be a torque acting on the object.
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$\begingroup$ I have very little idea about rigid body mechanics. Please, suggest me a good book to have clear conceptions. $\endgroup$ Commented May 8, 2017 at 8:39