Pulling a crate across the floor with a rope at an angle

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  • #1
jevans123
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Homework Statement:
I was told to be specific, so here it is:

Can you please help me answer this question that was given to me in the physics practice study guide? I am confused about this one and how to do so I would like someone to explain and walk me through how I should do this. Everybody keeps on telling me to reword this question and this is my second day just trying to ask a question on this platform. Can someone please help me out and I love you and I highly appreciate it if you did. Taking AP Physics 1 is quite challenging as a high schooler.

Roy pulls up at an angle on a rope attached to a crate to move the crate across a horizontal floor. The mass of the crate is 20.6 kilograms, Roy applies 115 newtons of force at an angle of 26.7 degrees, and the coefficient of kinetic friction between the crate and the floor is 0.349.

a. How hard does the floor push up on the crate?


b. What is the acceleration of the crate?
Relevant Equations:
f = mu x N
I drew out a motion diagram and used 115N sin(26.7 degrees) = 51.7 N but It turns out to be wrong.

Screen Shot 2022-10-13 at 9.41.41 PM.png
 
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  • #2
Orodruin
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Homework Statement:: Roy pulls up at an angle on a rope attached to a crate to move the crate across a horizontal floor. The mass of the crate is 20.6 kilograms, Roy applies 115 Newtons of force at an angle of 26.7 degrees, and the coefficient of kinetic friction between the crate and the floor is 0.349.

a. How hard does the floor push up on the crate?


b. What is the acceleration of the crate?
Relevant Equations:: f = mu x N

I drew out a motion diagram and used 115N sin(26.7 degrees) = 51.7 N but It turns out to be wrong.
You will have to be more specific than that in your attempts. You are not giving any reasoning or even stating what you are trying to compute.
 
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  • #3
jevans123
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I'm sorry, what does that even mean? Why are people just unwilling to help? Like I genuinely am asking a question and everybody just does some weird homework template and ohhh I haven't given enough detail and all. Like I am just asking for help and if you can't help then what is the point of commenting what you commented? I'm new so it would be nice to explain but I'm just so confused right now. What does "simultaneity" even mean and how is that even relevant?!
 
  • #4
jbriggs444
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What does "simultaneity" even mean and how is that even relevant?!
The word "simultaneity" appears in the signature that @Orodruin uses. It has nothing whatsoever to do with the response he provided.
a. How hard does the floor push up on the crate?
[...]
I drew out a motion diagram and used 115N sin(26.7 degrees) = 51.7 N but It turns out to be wrong.
Responding to the only work that was shown...

You have correctly calculated the vertical component of the tension force on the crate. Is that the quantity that was requested for part a?
 
  • #5
jevans123
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The word "simultaneity" appears in the signature that @Orodruin uses. It has nothing whatsoever to do with the response he provided.

[...]

Responding to the only work that was shown...

You have correctly calculated the vertical component of the tension force on the crate. Is that the quantity that was requested for part a?
Well, the reason why I calculated the vertical component was because it was asking for how hard does the floor push up on the crate which is essentially the normal force and the term "up" tells me to find the vertical component. I don't think that is all to it but its all that I have gotten and understood.
 
  • #6
jbriggs444
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Well, the reason why I calculated the vertical component was because it was asking for how hard does the floor push up on the crate which is essentially the normal force and the term "up" tells me to find the vertical component. I don't think that is all to it but its all that I have gotten and understood.
You are asked for the force of the floor on the crate.
You found the force of the rope on the crate.
Are those the same thing?

Have you drawn a free body diagram?
Can you make a list of the forces that act on the crate?
 
  • #7
jevans123
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List:

so we have,

the box being pulled at an angle of 26.7 degrees at 115 Newtons of force

then we have the weight of the crate going directly downward, Oh wait you made me think. Mass is not the same as weight. So now I'm thinking I use the mass of the box which is 20.6kg and multiply that by the Earth's constant acceleration of 9.80 and now the weight is 201.9 Newtons directly downward.

then we have the coefficient of kinetic friction directed the opposite of the way of the box at 0.349
 
  • #8
berkeman
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Have you drawn a free body diagram?
the box being pulled at an angle of 26.7 degrees at 115 Newtons of force
Please show us this (your FBD) before we go much farther. You can upload a PDF of JPEG copy of your diagram by using the "Attach files" link below the Edit window. Thank you.
 
  • #9
jevans123
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Screen Shot 2022-10-13 at 9.16.38 PM.png
 
  • #10
kuruman
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You don't show all forces acting on the block. The surface is also exerting a normal force and a force of kinetic friction.
 
  • #11
jevans123
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Screen Shot 2022-10-13 at 9.41.41 PM.png
 
  • #12
kuruman
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It is customary to draw the forces acting on the body as arrows with their tail on the body. See example below. Also, the force of friction is not 0.349. That number is the coefficient of kinetic friction. What is the expression that allows you to find the force of friction if you know the coefficient of kinetic friction?

FBD.png
 
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  • #13
jevans123
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OHHH ok so it would be this:

Ff=μkFN

But you need normal force to find that and I don't have normal force yet. Is normal force just the equal to the weight (in Newtons) of the object
 
  • #14
kuruman
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No. You need to add all the forces that have vertical components and set the sum equal to zero because the acceleration is zero in that direction.
 
  • #15
topsquark
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OHHH ok so it would be this:

Ff=μkFN

But you need normal force to find that and I don't have normal force yet. Is normal force just the equal to the weight (in Newtons) of the object
You seem to be missing a major concept in this problem. Newton's 2nd Law says that net force acting on an object is equal to its mass times its acceleration. ie. ##F_{net} = ma##. Force and acceleration are vectors so this is a vector law. So we have the net force component in the horizontal direction and the net force component in the vertical direction.

We know from the problem statement that the acceleration component in the vertical (call it y) direction is 0 because we are not lifting the box. We also know (or need to calculate) an acceleration in the horizontal (call it x) direction. So we have
##\sum F_x =ma##

##\sum F_y = 0##

Break each of your forces into components and plug them into the Newton's 2nd Law equations.

-Dan
 

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