What Is the Correct Expression for the Force Fo in an Inclined Plane Problem?

[wmrunner24]

Homework Statement

A block of mass m accelerates with acceleration g up a frictionless plane that is inclined at
an angle α above the horizontal. The force Fo that pushes the block is at an angle β above
the horizontal. Find the force Fo.

1. mg(1 + sin β)/cos(α + β)
2. mg(1 + sin α)/cos(α + β)
3. mg(1 + sin β)/cos(α − β)
4. mg(1 + sin α)/cos(α − β)
5. mg(1 − sin β)/cos(β − α)
6. mg(1 + sin α)/cos(β − α)
7. mg(sin β)/cos(α + β)
8. mg(1 + sin β)/cos(β − α)
9. mg(1 − sin β)/cos(α + β)
10. mg/cos(α + β)

Homework Equations

F=ma
a=g

The Attempt at a Solution

$$\Sigma$$F=ma=mg=Fu-Fg

where Fu is the force up the incline and Fg is the force of gravity (down)

Fg=mgsinα

Fu is a little more complicated to figure out. Here's my idea:

A is the adjacent side between the two triangles and Fu is the force parallel to the incline. A is also parallel to the horizontal. Because of this double parallel setup, it can be seen as two sets of parallel lines and transversals, so the new angle can be identifed as α using geometry. Thus:

cosβ=A/Fo
cosα=Fu/A

F$$_{u}$$=Fo*cosα*cosβ

So then:
mg=Fo*cosα*cosβ-mgsinα
mg(1+sinα)/(cosα*cosβ)=Fo

There is a trig identity that allows me to rewrite this as:

2mg(1+sinα)/(cos[α-β]+cos[α+β])=Fo

This is where I am stuck. I'm at an answer similar to those given but not quite there. Any help is greatly appreciated.

 

[issacnewton]

So then:
mg=Fo*cosα*cosβ-mgsinα

your upward force $$\vec{F}_u$$ is wrong. remember that a vector is same anywhere in the space as long as its direction and the length is the same. so bring that vector

$$\vec{F}_o$$

at the center of the block.. then what would be the angle between the vector

$$\vec{F}_o$$

and the inclined plane ?

 

[wmrunner24]

So...

cos(α+β)=Fu/Fo
Fo*cos(α+β)=Fu

So then:

mg=Fo*cos(α+β)-mgsin(α)
mg(1+sin(α))/cos(α+β)=Fo

So choice 2. Right?

 

[issacnewton]

yes........