Resultant of two forces in two different situations

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The discussion centers on the application of the law of cosines in vector addition, specifically addressing an error in the equation presented. The correct formula for vector addition is clarified as R² = P² + Q² + 2PQcosø, where ø is the angle between the vectors. Participants identify inconsistencies in the given values, noting that if P² + Q² = 100 and PQ = 69, it leads to an impossible negative result when applying the formula. The original poster acknowledges a correction regarding the values used in the calculations. Ultimately, the problem appears to be unsolvable based on the provided numbers.
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GiriBang said:
Root(P²+Q²+2PQcosø)
You have a sign error. Do a Google search for law of cosines.
 
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.
 
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Steve4Physics said:
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.
 
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kuruman said:
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.
 
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