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

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The discussion revolves around calculating the equilibrium force on a weight on a frictionless slope. A participant is struggling with part c of a homework problem, specifically determining the reading on a spring scale for a 5 kg weight on a 30-degree incline. The correct approach involves using the cosine formula to find the component of the weight acting along the slope, rather than incorrectly applying the sine function. Clarifications indicate that there is no horizontal force involved in this scenario. The conversation emphasizes the importance of understanding vector components in physics problems.
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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
 

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Welcome to PF!

Hi bettysuarez! Welcome to PF! :smile:

(I can't see the picture yet, but …)
bettysuarez said:
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 :confused: …

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

Just use the usual cosine formula for a component. :smile:
 
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!
 
bettysuarez said:
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! :smile:

Just use the cosine formula. :wink:​
 
Hint:

The horizontal weight is equal to the weight times the cosine of the angle of incline...
 
bleedblue1234 said:
Hint:

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

erm :redface:

… no such thing as horizontal weight! :confused:​
 

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