Crate on a slope - Angles and Tension

In summary, the conversation discusses a physics problem involving a crate being pushed up a slope with an attached mass hanging from a string. The goal is to find the angle at which the string makes with the lid of the crate during the push. The conversation includes equations such as F=ma and Newton's 2nd law, and suggests using a free body diagram to solve for the angle. The final answer is determined to be 78.1 degrees.
  • #1
mybrohshi5
365
0

Homework Statement



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.

Homework Equations



f=ma
Newton's 2nd law

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
 
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  • #2
What about a free body diagram with the pseudo forces included?
 
  • #3
Hi mybrohshi5! :wink:

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? :smile:
 
  • #4
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
Hi mybrohshi5! :smile:
mybrohshi5 said:
well the two forces would be the force due to gravity (its weight) and the normal force is that correct?

Normal force? Normal to what? It's not touching anything. :confused:

Add the two forces (one is unknown, so just give it a letter), and use F = ma.
 
  • #6
i thought there was a normal force but i guess cause it frictionless there is not a normal force?

im sorry I am so lost right now about adding the two forces (one is unknown) and use f = ma
 
  • #7
mybrohshi5 said:
i thought there was a normal force but i guess cause it frictionless there is not a normal force?

Are you talking about the crate?

I'm asking about the mass hanging, inside it.

What is the Ftotal = ma equation for the mass?
 
  • #8
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
mybrohshi5 said:
im just really confused... do i need to include tension in these two equations that i am trying to find?

You are confused, aren't you? :redface:

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

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

(and get some sleep! :zzz:)​
 
  • #10
im trying to figure out how this helps me solve the angle i need to find though...?
 
  • #11
are these the two equations i need:

mgsin(10) + T = ma

T - mgsin(theta) = ma
 
  • #12
mybrohshi5 said:
are these the two equations i need:

mgsin(10) + T = ma

T - mgsin(theta) = ma

uhh? what are they supposed to be? :confused:
 
  • #13
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
mybrohshi5 said:
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?

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
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
mybrohshi5 said:
my guess would be string/perpendicular to the string?

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
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 that's what i feel i am doing :(
 
  • #18
mybrohshi5 said:
so in the equation mg + T = ma

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

"replace"? :confused:

You use the components perpendicular to the string for the whole of mg + T = ma.
 
  • #19
im sorry i am completely lost.

thank you for your time and help though :)
 
  • #20
tiny-tim is there any other way you could maybe explain this to me a little differently? i don't seem to get it and its the only one on my homework i can't get :(
 
  • #21
Hi mybrohshi5! :smile:

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
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
mybrohshi5 said:
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.

What are you talking about?

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

There is only the tension in the string.
 
  • #24
Im sorry. I can't even think straight anymore. I am 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.
 

What is a crate on a slope?

A crate on a slope refers to a physical scenario in which a crate is placed on an inclined surface, creating a slope. This situation can affect the angles and tension experienced by the crate.

What is the significance of angles in a crate on a slope?

Angles play a crucial role in determining the stability and movement of a crate on a slope. The angle of the slope will affect the weight distribution of the crate and determine the direction and magnitude of the forces acting on it.

How does tension come into play in a crate on a slope?

Tension is the force that results from the pull or stretch between two objects. In a crate on a slope, the angle of the slope can affect the tension experienced by the crate, as well as the force needed to keep the crate from sliding down the slope.

What factors can affect the angles and tension of a crate on a slope?

The weight and shape of the crate, the angle of the slope, and the surface of the slope are all factors that can affect the angles and tension experienced by a crate on a slope. Other factors may include external forces such as wind or friction.

How can understanding angles and tension in a crate on a slope be useful?

Understanding the concepts of angles and tension in a crate on a slope can be useful in various practical situations, such as determining the safest angle to place a crate on a slope, calculating the amount of force needed to move a crate up or down a slope, or designing structures that can withstand slopes and inclines.

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