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Forces and Normal Force

  • Thread starter TS656577
  • Start date
[SOLVED] Forces and Normal Force

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

In the figure, a crate of mass m = 75 kg is pushed at a constant speed up a frictionless ramp (θ = 32°) by a horizontal force F. The positive direction of an x axis is up the ramp, and the positive direction of a y axis is perpendicular to the ramp. (a) What is the magnitude of F? (b) What is the magnitude of the normal force on the crate?

2. Relevant equations
F=ma


3. The attempt at a solution
After drawing a free body diagram...there is a movement to the right. F=mg which should mean F=75x9.8 which is 735N. I would think that the Normal Force would be found by F(g)cosx. Im fairly clueless about part A
 

rock.freak667

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When you resolve the weight you get mgsin32 down the plane and mgcos32 perpendicular to it right?

The only force acting in the +ve y direction is the the Normal reaction,R. So you know that answer.

Since the motion is up the plane the resultant force is

F-mgsin32=ma

What do you think the resultant force should be equal to while moving at a constant speed?
 
I am really confused with this
 
Wouldnt the Normal force be 632.32N? since the gravitational force is -632.32?
 

rock.freak667

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Wouldnt the Normal force be 632.32N? since the gravitational force is -632.32?
Yes it would.

Resultant force,ma=F-mgsin32

If you are moving with a constant speed up the ramp and your direction is not changing, what does that say about your acceleration?
 
there is no acceleration?
 

rock.freak667

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there is no acceleration?
Yes that is correct. If there is no acceleration then what is the resultant force acting on it?
 
Im sorry, but im really just not getting this one
 

rock.freak667

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Im sorry, but im really just not getting this one
ok then, we start over.

Do you agree that the force acting down the plane due to the horizontal component of the weight is mgsin(32) ?

A force,F, is pushing the crate up the plane. What is the resultant force here then?

F-mgsin(32) right?
 
That force would be 632.32 - 389.39?
 

rock.freak667

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That force would be 632.32 - 389.39?
Nope.


389.39 is the value of mgsin(32)

632.32 is the normal reaction. Those two forces are in different directions.


But anyhow.

The resultant force =F-389.39

If you are moving with a constant speed and your direction is not changing you aren't accelerating,right?
Then a=0 right?

Now if the resultant force is ma (Newton's 2nd law)

then ma=F-389.39
 
so it would equal 389.39 if acceleration is 0?
 
Well i put in 632.32 for the Normal force and its apparently wrong
 
I have no idea...i thought it was 632.32
 
Theonline program rejected it. Wasnt the other answer 389.39? because if it was, that was rejected as well
 
Re: [SOLVED] Forces and Normal Force

F=mg*tan(theta)
mg=mass*gravity
Fn=pithagren theorm
Fn=sqrtroot[(Fnx^2)+(Fny^2)]
Fny=mg
Fnx=F
 

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