# Weight down a slope

1. Oct 3, 2009

### bettysuarez

1. The problem statement, all variables and given/known data
I have attached a picture of the problem to this thread, I am having trouble with part c. I am getting an answer which is much larger than 24.5N

The systems shown in the figures are in equilibrium. If the spring scales are
calibrated in newtons, what do they read? (assume the incline in part (c) is frictionless.)

2. Relevant equations
Fw = mg

3. The attempt at a solution
I have attempted the solution and worked out the component of the weight force acting down the slope using Fw = mg / sin(30) = 49 / sin (30) = 98 N

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2. Oct 3, 2009

### tiny-tim

Welcome to PF!

Hi bettysuarez! Welcome to PF!

(I can't see the picture yet, but …)
You seem to finding the hypotenuse of a vector triangle

they don't work that way! (not without a horizontal force ).​

Just use the usual cosine formula for a component.

3. Oct 3, 2009

### bettysuarez

Sorry, I'm still a bit confused... The 5 kg weight is on a 30˚ slope and the spring scale is on the slope as well. How am I supposed to work out a horizontal force? Is that the force that will be recorded by the spring scale?

Thank you!

4. Oct 3, 2009

### tiny-tim

No, I was saying that there's no horizontal force!

Just use the cosine formula.

5. Oct 3, 2009

### bleedblue1234

Hint:

The horizontal weight is equal to the weight times the cosine of the angle of incline...

6. Oct 3, 2009

### tiny-tim

erm

… no such thing as horizontal weight!