Sliding down a ramp - quick question

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The discussion centers on the forces acting on a hamster sliding down a frictionless ramp, specifically the normal force exerted by the ramp on the hamster. It is established that while there is no force along the incline, the ramp exerts a normal force perpendicular to the incline. The normal force can be calculated using the hamster's weight and the angle of the ramp, with the formula N = mg cos(θ). The final scale reading, which accounts for both the ramp's weight and the vertical component of the normal force, is determined to be 9.47 N.

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If an object is sliding down a frictionless ramp will the force on the object by the ramp just be 0?

or will it be equal to the normal force?

Thanks :smile:
 
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mybrohshi5 said:
If an object is sliding down a frictionless ramp will the force on the object by the ramp just be 0?

or will it be equal to the normal force?

Thanks :smile:
Thre will be no force by the ramp on the object along the incline, but perpendicular to the incline, yes, the ramp will exert a normal force on it.
 
mybrohshi5 said:
So for this case...

http://session.masteringphysics.com/problemAsset/1000050976/7/07.CP53.jpg

While the hamster is sliding down, what is the magnitude of the force on the hamster by the block? The block is frictionless.

would it just be the normal force of the hamster which is perpendicular to the block?
The normal force of the block on the hamster would be equal and opposite to the normal force of the hamster on the block, per Newton's 3rd law... It's magnitude must be determined using trig and one of the other of Newton's laws.
 
So would the magnitude of the force on the hamster by the block be...

x = F_g and F_g = .25(9.8) = 2.45 N

so the normal force would be

Cos(40) = x/n

n=2.45/cos(40) = 3.198 N

HMMM this is wrong though.
 
What's the mass or weight of the hamster??
 
sorry i forgot to include that.

250g hamster so 0.25 kg

820g ramp so 0.820 kg
 
If you look carefully at the geometry and Newton's first law, the normal force of the block on the hamster must equal the hamster's weight times cos 40 (N=mgcos40). Convince yourself that this is true.
 
thank you. that was correct. i need to work on taking more time when drawing my diagrams haha.
 
  • #10
I cannot get this last part.

For the situation described in Part C, what does the scale read as the hamster slides down?

part C is the one i just figured out.

I thought the scale would just read the force the hamster has on the ramp 1.88 N plus the weight of the ramp 8.04 N

so 9.92 N but that is wrong

so i thought maybe its just the total weight (.25+.82)(9.8)=10.5 N but that was wrong as well.

I don't know what to do.
 
  • #11
Only part of the hamster's weight is measured on the scale. Part is directed down the ramp which doesn't measure on the scale but the other part pushes the ramp down and is measured on the scale.
 
  • #12
Jebus_Chris said:
Only part of the hamster's weight is measured on the scale. Part is directed down the ramp which doesn't measure on the scale but the other part pushes the ramp down and is measured on the scale.

I tried finding the weight with using the normal force of the hamster and using the weight of the hamster and both of those are wrong so my only hamster component left would be the one parallel to the ramp?

Is that the one i add to the weight of the ramp?
 
  • #13
You may have to take the y component of the normal force.
 
  • #14
isnt the y component of the normal force just the same as the weight due to gravity?

or are you considering the y component to be parallel to the ramp?
 
  • #15
so the answer ended up being 9.47

Im not sure how its exactly that because i got 9.6 for my answer.

i added the weight of the block plus the component force of the hamster that is parallel to the slope of the ramp so

9.8(.82) + (.25)(9.8)sin(40) = 9.6

anyone know why its 9.47

did i do it right and the online homework just rounded differently or something like it usually does?

thanks
 
  • #16
mybrohshi5 said:
so the answer ended up being 9.47

Im not sure how its exactly that because i got 9.6 for my answer.

i added the weight of the block plus the component force of the hamster that is parallel to the slope of the ramp so

9.8(.82) + (.25)(9.8)sin(40) = 9.6

anyone know why its 9.47

did i do it right and the online homework just rounded differently or something like it usually does?

thanks
The block weight and parallel comp of the hamster act in different directions, so adding them together doesn't give you anything at all to work with.. You should be adding the block weight and the vert comp of the Normal force between the hamster and ramp to solve for the scale reading.. The key is to find the vert comp of the normal force between the ramp and hamster. The hamster is accelerating in the y direction, so you must use Newton 2 when looking at the forces acting on the hamster in the y direction. It's a bit of trig involved,and a bit tricky.
 
  • #17
yeah that sounds like a lot of work. thank you for explaining all that :)
 

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