Question

Two forces P,Q act at a point along two straight line making an angle alpha with each other and R is the their resultant two other forces P'Q' acting along the same line has a resultant R' prove that- the angle between the lines of action of resultant is -
Cos^-1{PP'+QQ'+ cos alpha (PQ'+P'Q)}\RR'

Answer 1

The angle between the lines of action of the resultants R and R' is $Cos^-1\frac{{PP'+QQ'+ cos \alpha (PQ'+P'Q)}}{RR'}$.

Given:

Two forces P and Q act at a point along two straight lines making an angle alpha with each other and R is their resultant. Two other forces P' and Q' act along the same lines as P and Q has a resultant R'.

To Find:

the angle between the lines of action of resultants R and R'.

Solution:

We can provide the solution to this problem in the following way.

(Parallelogram law helps to find the magnitude and the direction of the resultant when two forces act at a point along the adjacent sides of a parallelogram.)

Let us suppose that the resultant R makes an angle $\theta$ with the force P or P' and the resultant R' makes an angle $\phi$ with the force P or P'.

We can write the magnitudes of R and R' in the following way.

$R=\sqrt{P^{2} +Q^{2} +2PQcos\alpha }$

$R'=\sqrt{P^{'2} +Q^{'2} +2P'Q'cos\alpha }$

We can evaluate the following.

$RR'=\sqrt{(P^{2} +Q^{2} +2PQcos\alpha )(P^{'2} +Q^{'2} +2P'Q'cos\alpha }})\\RR'=\sqrt{(PP'+QQ'+cos\alpha (P'Q+PQ')^{2}+(P'Qsin\alpha +PQ'sin\alpha )^{2} }$

We can write the directions of R and R' in the following way.

$tan\theta =\frac{Qsin\alpha }{P+Qcos\alpha } \\tan\phi =\frac{Q'sin\alpha }{P'+Q'cos\alpha }$

We can find the angle between the resultants R and R' in the following way

$tan(\theta -\phi )=\frac{tan\theta -tan\phi }{1-tan\theta tan\phi } \\=\frac{\frac{Qsin\alpha }{P+Qcos\alpha }-\frac{Q'sin\alpha }{P'+Q'cos\alpha }}{1+\frac{Qsin\alpha }{P+Qcos\alpha }\frac{Q'sin\alpha }{P'+Q'cos\alpha }}\\=\frac{P'Qsin\alpha -PQ'sin\alpha }{PP'+QQ'+cos\alpha (P'Q+PQ')}$

We can use the trigonometric identity $sec^{2} A= tan^{2} A+1$ to deduce the following.

$cos(\theta -\phi )= \frac{{PP'+QQ'+ cos \alpha (PQ'+P'Q)}}{RR'}\\(\theta -\phi )=Cos^-1\frac{{PP'+QQ'+ cos \alpha (PQ'+P'Q)}}{RR'}$

Therefore, we have proved that the angle between the lines of action of the resultants R and R' is $Cos^-1\frac{{PP'+QQ'+ cos \alpha (PQ'+P'Q)}}{RR'}$.

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