What is the Tension on a Spring Scale for an Object on an Incline?

[saber1357]

Homework Statement

I am trying to understand how these Newton's laws work, specifically with an object on an incline.
If a 5.00kg object is attached to a Newton spring scale on an incline (incline makes 30 degrees with the ground), what reading is the scale giving?

Homework Equations

Fx = m*g*sin(theta)
Fy = n - m*g*cos(theta)

The Attempt at a Solution

Since the object is not accelerating upward, the netforce will equal zero. And I can solve for the force in the x direction. But how can I use these numbers to find the tension on the spring scale?

 

[learningphysics]

There are 2 forces in the x-direction... mgsin(theta) and the force the spring exerts (and this is what the spring scale reads)...

the two forces balance each other.

 

[saber1357]

So I can simply ignore everything else, solve for the force in the x direction and what the spring reads is opposite of that?

 

[learningphysics]

So I can simply ignore everything else, solve for the force in the x direction and what the spring reads is opposite of that?

yes, I believe so. The object and the spring scale are both on the incline right? If that's the case then the spring scale simply reads mgsin(theta).

is the spring scale attached to the incline?

 

[saber1357]

It's not attached to the incline, but a wall directly behind it. I believe that if the spring scale wasn't attached to anything, then there would be no purpose for it.

 

[learningphysics]

It's not attached to the incline, but a wall directly behind it. I believe that if the spring scale wasn't attached to anything, then there would be no purpose for it.

yes, that's true.