Concurrent Forces Problem: Finding Unknown Forces and Angles

[lachy874125]

Homework Statement

Two forces acting at the origin of the x-y axes have a resultant of 2000 N in the positive direction of y. If one force acts at 40° to the x direction and the other has a magnitude of 1800 N, find:
(a) The magnitude of the 40° force.
(b) The direction of the 1800 N force.

Homework Equations

The Attempt at a Solution

I drew a free-body diagram but still can't find any other values to help solve the problem.

 

[FermiAged]

Provide your free-body diagram as a start. That way we can see where you are at and have a common reference. The problem has all the info needed but you will have to use a trial and error solution method.

 

[donpacino]

You don't need to use a trial and error solution method...
the sine of the angle relative to the x-axis will tell you the y component of that vector.
therefore...

y=A*sin(theta)

we know the two forces add up to 2 kN, so
y1+y2=2 kN
y1=A*sin(40)
y2=1800*sin(theta)

there are two unknowns with one equation. therefore there are an infinite number of solutions, unless you further constrain the answer

 

[SteamKing]

You don't need to use a trial and error solution method...
the sine of the angle relative to the x-axis will tell you the y component of that vector.
therefore...

y=A*sin(theta)

we know the two forces add up to 2 kN, so
y1+y2=2 kN
y1=A*sin(40)
y2=1800*sin(theta)

there are two unknowns with one equation. therefore there are an infinite number of solutions, unless you further constrain the answer

Not so fast. You've stopped short in your analysis. Remember, you still have horizontal components to consider.

If you continue your analysis, you may stumble across a basic trigonometric identity which will prove helpful in eliminating most of that infinite number of solutions.

 

[donpacino]

Not so fast. You've stopped short in your analysis. Remember, you still have horizontal components to consider.

If you continue your analysis, you may stumble across a basic trigonometric identity which will prove helpful in eliminating most of that infinite number of solutions.

Haha. I'm a dummy. I interpreted op as saying there was an unknown x force, not as the force was purely y.

@OP. using a method similar to the one i previously used, you can determine an equation for net x force. If the force is purely in the y direction that means the x force is...
using the resulting equation you have 2 equations, 2 unknowns, and you can solve for the answer

 

[FermiAged]

The additional constraint is that the sum of forces in the x direction is zero. That gives two equations with two unknowns which contain trigonometric functions. I used EXCEL to iterate on angle to see where the forces were equal. Before doing that, you could try assuming a small angle and letting sin(theta) = theta and cos(theta) = 1-theta. The equations can then be solved for F and theta. Plug them back into the original equations to see if the answers are consistent.

 

[SteamKing]

The additional constraint is that the sum of forces in the x direction is zero. That gives two equations with two unknowns which contain trigonometric functions. I used EXCEL to iterate on angle to see where the forces were equal. Before doing that, you could try assuming a small angle and letting sin(theta) = theta and cos(theta) = 1-theta. The equations can then be solved for F and theta. Plug them back into the original equations to see if the answers are consistent.

You don't need to iterate. By analyzing the relationships between the horizontal and vertical components, you'll eventually stumble on the identity sin^2 + cos^2 = 1. By suitable algebraic manipulations, you'll wind up with a quadratic equation in the unknown magnitude of one of the forces, which can then be used to determine the unknown angle of the other force vector.