 Hello and welcome to the session. Let's work out the following problem. It says to like parallel forces p and q Act on a rigid body at a and b respectively if p and q are Interchanged in position show that the point of application of the resultant will be displaced Through a distance p minus q upon p plus q into a b. So let's now move on to the solution and let The forces a and b respectively on Now let p a point on a b through which The resultant passes So we are given two light parallel forces p and q acting at a and b and Their resultant acts through a point c which is on a b Now by the result of light parallel forces we have p upon cb is Equal to q upon a c is equal to p plus q upon a b Now using these two we have a c is equal to q into a b upon p plus q now when the forces are Interchanged in position Let's see that Be the point through which the resultant passes so if the Forces interchange in position Then resultant acts through the point c dash then again by the theorem result of light parallel forces will have a C dash is equal to p into a b upon p plus q as we discussed above right now We have to find the distance through which the resultant is displaced. So we have to find the distance cc dash Which is given by a c dash Minus a c right Now a c dash is p into a b upon p plus q and a c is q into a b upon p plus q So this becomes Taking a mine a b common we have p minus q into a b upon p plus q Hence we have proved that the resultant is displaced through a distance p minus q upon p plus q into a b So this completes the question and the session by for now take care. Have a good day