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A particle is in equilibrium (a force vector problem) please help!!

A particle is in equilibrium under the action of three horizontal forces P, Q and R. The magnitudes of P, Q and R are 25 N, Q N, and R N respectively, and the cosine of the angle between P and Q is -0.96. Show that R^2 = (Q - 24)^2 + 7^2.

Hence find:
(a) the least possible value of R,
(b) the corresponding angle between Q and R.:confused:
Reply 1
Avatar for kanojyoxx
kanojyoxx
12
Do you have a diagram?
Reply 2
Avatar for lasreedhar
lasreedhar
OP
The diagram was not given in the book. But this is what i did.

My working:
R^2 = P^2 + Q^2 - 2(P)(Q) Cos (theta)
R^2 = 25^2 + Q^2 - 2(25) (Q) (-0.96)
R^2= 25^2 + Q^2 + 48 Q
R^2 = Q^2 + 48Q + 25^2
R^2= (Q)^2 + 2(Q)24 + (24)^2 - (24)^2 + 625
R^2= (Q + 24)^2 -576 + 625
R^2 = (Q + 24)^2 + 49
R^2 = (Q + 24)^2 + 7^2
it looks like i am almost there. But why is it (Q-24) in the required proof
and not (Q + 24) as i have got in my proof.
Am i making some mistake? please suggest. Thanks
Reply 3
Avatar for lasreedhar
lasreedhar
OP
Original post by kanojyoxx
Do you have a diagram?


Can You kindly help me? I am stuck!!

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