Seemingly simple system of equations

[ElTaco]

Homework Statement

There are two forces P and Q that are applied to a crate, with their respective magnitudes 100 and 200 N. One is applied at an angle upward with angle theta, and one is applied at an angle downward with angle phi. Both have positive horizontal components and the sum of the two forces has a magnitude of 250 N directly horizontal to the right. I need to find the angles theta and phi.

Homework Equations

None explicitly given.

The Attempt at a Solution

100sin(theta) = 200sin(phi)
100cos(theta) + 200cos(phi) = 250
theta = arcsin(2sin(phi))
Then I plug it in and cannot solve.

Another route I tried was keeping P and Q in instead of substituting in their magnitudes.
Psin(theta) = Qsin(phi)
Pcos(theta) + Qcos(phi) = 250
P = Qsin(phi)/sin(theta)
Qsin(phi)/tan(theta) + Qcos(phi) = 250
Again I cannot finish solving.

If something was badly worded or confusing please let me know. This problem is from a book so rather than writing the problem as exactly stated, I had to add some more information because there was a picture as well. Thanks for your time, I'm really stuck on this.

 

[chronos98]

I'm assuming you at least have a graphing calculator. In function mode just do y1=250 and y2=100cos(arcsin(2sin(phi))) + 200cos(phi) = 250, which is what you had. Find the right range and domain and the intersection points will be your answer. you could also set it equal to 0 and fin the zeros.

 

[gneill]

Your two equations look like:

$$P cos(\theta) + Q cos(\phi) = F_h$$
$$P sin(\theta) + Q sin(\phi) = 0$$

Isolate cos(Θ) and sin(Θ), or cos(Φ) and sin(Φ) on one side of each equation. Square both sides of each. Add the equations. sin² + cos² = 1. Solve for the remaining angle. Note that sign information may be lost in the squaring process, so check the result for signs.

 

[chronos98]

His solution is more intelligent. I recommend that.

 

[rwisz]

I would approach this problem by first identifying the known and unknown variables and then using the given equations to solve for the unknown angles theta and phi.

Known variables:
- Magnitudes of forces P and Q (100 N and 200 N)
- Horizontal component of the resultant force (250 N)
- Relationships between the forces and angles (Psin(theta) = Qsin(phi) and Pcos(theta) + Qcos(phi) = 250)

Unknown variables:
- Angles theta and phi

Using the given equations, we can set up a system of equations and solve for the unknown angles:
Psin(theta) = Qsin(phi)
Pcos(theta) + Qcos(phi) = 250

We can substitute the known values for P and Q:
100sin(theta) = 200sin(phi)
100cos(theta) + 200cos(phi) = 250

We can then use trigonometric identities to simplify these equations:
sin(theta) = 2sin(phi)
cos(theta) + 2cos(phi) = 5

We can then use the Pythagorean identity (sin^2(x) + cos^2(x) = 1) to solve for the unknown angles:
sin^2(theta) = 4sin^2(phi)
cos^2(theta) + 4cos^2(phi) = 25

Subtracting the first equation from the second, we get:
cos^2(theta) = 25 - 4sin^2(phi)

We can then substitute this into the first equation and solve for sin(theta):
sin^2(theta) = 4sin^2(phi)
sin^2(theta) = 4(1 - cos^2(theta))
sin^2(theta) = 4 - 4cos^2(theta)
sin(theta) = sqrt(4 - 4cos^2(theta))

We can then substitute this into the first equation and solve for phi:
sin(theta) = 2sin(phi)
sqrt(4 - 4cos^2(theta)) = 2sin(phi)
4 - 4cos^2(theta) = 4sin^2(phi)
cos^2(theta) = 1 - sin^2(theta)
cos(theta) = sqrt(1 - sin^2(theta))

Using these equations, we can solve for the unknown angles theta and phi.

Overall, this problem involves using trigonometric identities and equations to solve for the unknown angles