Homework Help: Crate on a slope - Angles and Tension

1. Feb 14, 2010

mybrohshi5

1. The problem statement, all variables and given/known data

A large empty crate, with its lid in place, has a 9.50 kg mass hanging from a string attached to the center of the lid. If the crate were sitting (at rest) on a flat surface, the string would simply hang straight down such that the mass would not be touching the floor of the crate.

The crate is now pushed up a frictionless 10.0 degree slope with an acceleration of 2.25 m/s^2. During the push, what angle does the string make with the lid of the crate? Hint: you will need to solve two equations in two unknowns.

2. Relevant equations

f=ma
newton's 2nd law

3. The attempt at a solution

I found the force required to push it up the ramp

f= 9.8(9.5)sin10 = 16.2N

Not sure if i even need that but i did it anyways.

so i really have no idea what 2 equations i need to get in order to solve the angle of the string?

can anyone help me on this please.

thank you

2. Feb 14, 2010

aim1732

What about a free body diagram with the pseudo forces included?

3. Feb 14, 2010

tiny-tim

Hi mybrohshi5!

The acceleration of the mass will be the same as that of the crate (and in the same direction). And there are two forces on the mass.

So use good ol' Newton's second law on the mass …

what do you get?

4. Feb 14, 2010

mybrohshi5

well the two forces would be the force due to gravity (its weight) and the normal force is that correct?

im a little confused about what i need to find from what you just stated above.

thank you for any further explanations

5. Feb 14, 2010

tiny-tim

Hi mybrohshi5!
Normal force? Normal to what? It's not touching anything.

Add the two forces (one is unknown, so just give it a letter), and use F = ma.

6. Feb 14, 2010

mybrohshi5

i thought there was a normal force but i guess cause it frictionless there is not a normal force?

im sorry im so lost right now about adding the two forces (one is unknown) and use f = ma

7. Feb 14, 2010

tiny-tim

Are you talking about the crate?

What is the Ftotal = ma equation for the mass?

8. Feb 14, 2010

mybrohshi5

sorry i was reading the question completely wrong

for some reason i am having a really hard time wrapping my head around what is going on still though and how to find these equations :(

the mass will have two forces on it correct.

one force will be the weight

and the other force will be the force parallel to the ramp just in the opposite direction?

im just really confused.... do i need to include tension in these two equations that i am trying to find?

9. Feb 14, 2010

tiny-tim

You are confused, aren't you?

Yes, the tension is the only other force on the mass.

ok, so solve the vector equation: mg + T = ma.

(and get some sleep! :zzz:)​

10. Feb 14, 2010

mybrohshi5

im trying to figure out how this helps me solve the angle i need to find though....?

11. Feb 14, 2010

mybrohshi5

are these the two equations i need:

mgsin(10) + T = ma

T - mgsin(theta) = ma

12. Feb 14, 2010

tiny-tim

uhh? what are they supposed to be?

13. Feb 14, 2010

mybrohshi5

goodness i am not exactly sure. i am so confused on how to get two equations for this.

ok, so solve the vector equation: mg + T = ma.

when you said this should i solve it for the tension?

14. Feb 14, 2010

tiny-tim

Yes. T has magnitude and direction (and the question only asks you for the direction).

You will need to take components in a suitable direction, of both a and the forces.

You have a choice of three pairs of components:

horizontal/vertical

slope/normal

string/perpendicular-to-string …

which do you think will be quickest?

15. Feb 14, 2010

mybrohshi5

my guess would be string/perpendicular to the string?

but maybe would it be slope/normal because the angle i am trying to find has to deal with the slope and the crate....

16. Feb 14, 2010

tiny-tim

Yes, that will be quicker because you don't want to find the magnitude of T, so you can avoid it by using components perpendicular to it.

17. Feb 14, 2010

mybrohshi5

so in the equation mg + T = ma

do i replace T with the components perpendicular to the string?

gosh i feel so stupid right now. i cannot even begin to explain how confused i am about this whole problem :(

am i just making it much more difficult than it actually is because thats what i feel i am doing :(

18. Feb 14, 2010

tiny-tim

"replace"?

You use the components perpendicular to the string for the whole of mg + T = ma.

19. Feb 14, 2010

mybrohshi5

im sorry i am completely lost.

thank you for your time and help though :)

20. Feb 15, 2010

mybrohshi5

tiny-tim is there any other way you could maybe explain this to me a little differently? i dont seem to get it and its the only one on my homework i cant get :(

21. Feb 15, 2010

tiny-tim

Hi mybrohshi5!

I'm sorry, on this forum we don't do the problem for you.

What equations have you got so far (with the numbers in)?

22. Feb 15, 2010

mybrohshi5

I know that F=ma

the sum of the y forces on the block attached to the string will be gravity which will be straight down and the normal force which will be perpendicular to the surface of the block attached to the string.

the sum of the x forces will be the force parallel to the slope of the ramp so 9.8(9.5)sin10

and i know that mg + T = ma like you stated before.

is that correct so far?

im really trying to understand this.

im just not sure how to use all of this to get to the 2 equations i need.

23. Feb 15, 2010

tiny-tim

There is no relevant surface. There is no normal force (I've said that before).

There is only the tension in the string.

24. Feb 15, 2010

mybrohshi5

Im sorry. I cant even think straight anymore. im so stressed out. i have a physics exam tomorrow and this homework assignment due tonight.

Thanks for trying to help me. sorry i was such a pain.