Resultant of two forces in two different situations

[GiriBang]

Homework Statement

Determine the magnitude of the two forces such that if they act at right angles, their
resultant is 10N. But if they act at 60° their resultant is 13N.

Relevant Equations

Root(P²+Q²+2PQcosø)

https://cdn.discordapp.com/attachments/843776584255406080/847500756194033754/20210527_211342.jpg

 

[Mister T]

Root(P²+Q²+2PQcosø)

You have a sign error. Do a Google search for law of cosines.

 

[Steve4Physics]

Note that "Root(P²+Q²+2PQcosø)" is not a equation.

EDIT: - correction: in the text below, I originally typed '2PQ' instead of 'PQ'. I have corrected this.

You have worked out that P² + Q² =100 and PQ = 69.
Finding P and Q is now just an algebra problem. I’ll start you off...

P =√(100 – Q²) so that PQ = 69 becomes:
√(100 - Q²)Q = 69

Can you complete it from there?

@Mister T, the law of cosines for the sides of a triangle is R² =P² +Q² - 2PQcosθ. But the rules for vector-addition mean that we need the (closely related) formula R² = P² +Q² + 2PQcosø where ø is the angle between vectors P and Q. If you draw the vector addition triangle or parallogram, and remember cosθ = -cos(180º- θ), you should see what is going on.

 

[kuruman]

You have worked out that P² + Q² =100 and 2PQ = 69

OP has worked out that
P² + Q² =100 and PQ = 69.

I agree that P² + Q² =100 and that PQ = 69 which makes 2PQ = 138. But if that were true, then P² + Q² - 2PQ = (P-Q)² = 100 - 138 = -38. The square of a (real) number is never negative.

It looks like the numbers are bad.

 

[Steve4Physics]

OP has worked out that
P² + Q² =100 and PQ = 69.

I agree that P² + Q² =100 and that PQ = 69 which makes 2PQ = 138. But if that were true, then P² + Q² - 2PQ = (P-Q)² = 100 - 138 = -38. The square of a (real) number is never negative.

It looks like the numbers are bad.

The question is insoluble - well spotted. It might be a mistake in the question but it could be deliberate.

I've corrected my mistake in Post #3 (should have said PQ = 69, not 2PQ = 69). Thanks.