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Normal force of a block with a massless rope

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Homework Statement


Mass of block: 19kg
Applied force: 225N at 16 degrees to accelerate block from rest
Coefficient of static: 0.55
Coefficient of kinetic: 0.30


Homework Equations


Unknown

The Attempt at a Solution


I have solved for parallel tension and perpendicular tension

Tperp = sin(16)225N = 62.0N
Tpar = cos(16)225N = 216.3N

I am unsure if I need these values, but I need to solve for the normal force of the block
 

Answers and Replies

  • #2
phinds
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To my mind, discussing such problems without a vector diagram (free body diagram) of what's going on is a total waste of time.
 
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  • #3
CWatters
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I am unsure if I need these values, but I need to solve for the normal force of the block
You will need them but I agree with phinds.

You can use MS Paint or any free drawing program for FBD diagrams. Doesn't need to be a work of art.
 
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sorry for that, had to change the file type. thank you for the persistance
 

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haruspex
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sorry for that, had to change the file type. thank you for the persistance
The image is tiny. If I expand it to a decent size the text is illegible.
Please quote the question exactly as given to you. If you cannot find a way to upload a reasonable diagram then provide a clear description of the set-up.
 
  • #8
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Natasha uses a massless rope to pull a 19kg box of books, which is initially at rest, across a rough floor by applying a 225N force at 16 degrees above the horizontal. the coefficients of friction between the box and floor are μs :0.55 and μk : 0.30. What is the magnitude of the normal force by the floor on the box of books? After finding the Normal Force, find the magnitude of the box's acceleration as it begins to move from rest.

the vector NF,B is directed straight up, with the vector WE,B straight down. a small force of FkT,Bis aimed straight to the left, while the Tension force of TR,B is to right at an angle of 16 degrees.
 
  • #9
CWatters
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Ok so in the OP you correctly calculated the vertical (Tperp) and horizontal (Tpar) components of the applied force. Now can you write equation for the total vertical force on the box?
 
  • #10
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Im not sure what the equation is.
 
  • #11
haruspex
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Im not sure what the equation is.
The usual one, ΣF=ma.
Is there any vertical acceleration?
 
  • #12
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The usual one, ΣF=ma.
Is there any vertical acceleration?
I do not believe so.
 
  • #13
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I solved this problem, thanks to all for the help.
The answer was 124.2 newtons
 
  • #14
haruspex
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I solved this problem, thanks to all for the help.
The answer was 124.2 newtons
Right.
 

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