How do you find the angle...?

Question

Q: Find the able between the forces given the magnitude of their resultant. (Hint: Write Force 1 as a vector in the direction of the positive x-axis, and Force 2 as a vector at an angle theta with the positive x-axis.)
Force 1 = 3000 pounds
Force 2 = 1000 pounds
Resultant Force = 3750 pounds

How do you solve this problem? Would you write Force 1 and Force 2 in a component form, like <3000, 1000>?

Answer 1

Use the hint. But to make it easier, divide everything by 1000 first. That doesn't change the problem at all as long as you do it everywhere.

F1 = <3000/1000, 0> = <3, 0>

F2 = <1000/1000 cos θ, 1000/1000 sin θ> =

F1 + F2 = <3 + cos θ, sin θ>

||F1 + F2|| = √((3 + cos θ)² + (sin θ)²) = 3750/1000 = 3.75

Square both sides, and simplify:

9 + cos² θ + 6 cos θ + sin² θ = 14.0625

Group the sin² θ + cos² θ together:

9 + (sin² θ + cos² θ) + 6 cos θ = 14.0625

Substitute 1 for it:

9 + 1 + 6 cos θ = 14.0625

10 + 6 cos θ = 14.0625

Subtract 10 from both sides:

6 cos θ = 4.0625

Divide by 6:

cos θ = 4.0625/6 = 0.67708

Take the arccos:

θ = 47.38°

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Pretty sure my answer is the right one.

To check:

F1 = <3000, 0>
F2 = <1000 cos(47.38°), 1000 sin(47.38°)> = <677.0833, 735.9064>

F1 + F2 = <3677.0833, 735.9064>

||F1 + F2|| = √(3677.0833² + 735.9064²) = 3750...check

Answer 2

Given that the two forces are at an angle, θ to each other - you need to use the *parallelogram law of vector addition (one vector lies between the z- and x-axis). So the resultant is quickly calculated by using the cosine law r^2 = F1^2 + F2^2 - 2(F1)(F2) cos R. Once R is solved for the angle is 73.47 degrees. .. check the source if you have any discrepancies

Answer 3

51.34 degrees
tangent of angle= (3750/3000)

Answer 4

oh. I see you use law of cosines.
c^2=a^2+b^2-2ab cos C
...
(dont be scared of big numbers!)
...
the ans is 132.6161427 degrees.

Answer 5

you lost me at "Q"