Forces - Algebra Question?

The magnitude of resultant R of two forces P and Q acting at an angle θ is given as

R²=P²+Q²+2PQ cosθ

When P= 30 Ponds Q= 35 Pounds and R= 42 Pounds, we have

42²=30²+35²+2×30×35cosθ

→1764=900+1225+2100cosθ

→2100cosθ= −361

cosθ=−361/2100

and θ=99.90°

Use the parallelogram of forces.

Draw a parallelogram with sides 30 and 35 and with diagonal 42.

Then use the cosine formula, a^2 = b^2 + c^2 - 2bccosC

There angle between the vectors is divided by the diagonal.

First angle is given by

35^2 = 42^2 + 30^2 - 2*30*42*cosC

1225 - 1764 - 900 = - 2520cosC

cosC = 0.57103

C = 55.18 degrees

Second angle is given by

30^2 = 35^2 + 42^2 = 2*35*42cos B

900 - 1225 - 1764 - 2940cosB

cosB = 0.71054

B = 44.72 degrees

Angle between vectors = 55.18 + 44.72 = 99.9 degrees

So let the angle between the two forces be theta. Then the magnitude of the resulting force is sqrt(F1^2 + F2^2 + 2*F1*F2 cos(theta))

So F1 = 30 pounds and F2 = 35 pounds

sqrt(30^2 + 35^2 + 2*30*35*cos(theta)) = 42

sqrt(900 + 1225 + 2100 cos(theta)) = 42

2125 + 2100cos(theta)= 1764

2100cos(theta) = -361

cos(theta) = -0.1719

so theta = 99.898 (degrees) = 100 degrees

R^2 --> P^2 + Q^2 - 2PQ cos x

ie 42^2 ---> 30^2 +35^2 -2*30*35 cosx

ie cosx ---> (30^2 +35^2 -42^2)/(2*30*35) ---> 0.1719 giving x as 80.1 deg