What is the equilibrium force on a weight down a slope?

[bettysuarez]

Homework Statement

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.)

Homework Equations

Fw = mg

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

 

[tiny-tim]

Welcome to PF!

Hi bettysuarez! Welcome to PF!

(I can't see the picture yet, but …)

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

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.

 

[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!

 

[tiny-tim]

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?

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

Just use the cosine formula. ​

 

[bleedblue1234]

Hint:

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

 

[tiny-tim]

Hint:

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

erm

… no such thing as horizontal weight! ​