# Sliding down a ramp - quick question!

1. Feb 13, 2010

### mybrohshi5

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

2. Feb 13, 2010

### PhanthomJay

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.

3. Feb 13, 2010

### mybrohshi5

4. Feb 13, 2010

### PhanthomJay

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.

5. Feb 13, 2010

### mybrohshi5

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.

6. Feb 13, 2010

### PhanthomJay

What's the mass or weight of the hamster??

7. Feb 13, 2010

### mybrohshi5

sorry i forgot to include that.

250g hamster so 0.25 kg

820g ramp so 0.820 kg

8. Feb 13, 2010

### PhanthomJay

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.

9. Feb 13, 2010

### mybrohshi5

thank you. that was correct. i need to work on taking more time when drawing my diagrams haha.

10. Feb 13, 2010

### mybrohshi5

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 dont know what to do.

11. Feb 13, 2010

### Jebus_Chris

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. Feb 13, 2010

### mybrohshi5

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. Feb 13, 2010

### Jebus_Chris

You may have to take the y component of the normal force.

14. Feb 13, 2010

### mybrohshi5

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. Feb 14, 2010

### mybrohshi5

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 hw just rounded differently or something like it usually does?

thanks

16. Feb 14, 2010

### PhanthomJay

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 betwen 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. Feb 14, 2010

### mybrohshi5

yeah that sounds like a lot of work. thank you for explaining all that :)

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