Forces and Normal Force

  • Thread starter TS656577
  • Start date
  • #1
62
0
[SOLVED] Forces and Normal Force

Homework Statement



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?

Homework Equations


F=ma


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
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31
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?
 
  • #3
62
0
I am really confused with this
 
  • #4
62
0
Wouldnt the Normal force be 632.32N? since the gravitational force is -632.32?
 
  • #5
rock.freak667
Homework Helper
6,223
31
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?
 
  • #6
62
0
there is no acceleration?
 
  • #7
rock.freak667
Homework Helper
6,223
31
there is no acceleration?

Yes that is correct. If there is no acceleration then what is the resultant force acting on it?
 
  • #8
62
0
Im sorry, but im really just not getting this one
 
  • #9
rock.freak667
Homework Helper
6,223
31
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?
 
  • #10
62
0
That force would be 632.32 - 389.39?
 
  • #11
rock.freak667
Homework Helper
6,223
31
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
 
  • #12
62
0
so it would equal 389.39 if acceleration is 0?
 
  • #13
rock.freak667
Homework Helper
6,223
31
so it would equal 389.39 if acceleration is 0?
It would be equal to that because the crate is not accelerating.
 
  • #14
62
0
Well i put in 632.32 for the Normal force and its apparently wrong
 
  • #15
rock.freak667
Homework Helper
6,223
31
Well i put in 632.32 for the Normal force and its apparently wrong

What is the answer?
 
  • #16
62
0
I have no idea...i thought it was 632.32
 
  • #17
rock.freak667
Homework Helper
6,223
31
I have no idea...i thought it was 632.32

Then how do you know the answer is not 623.95N?
 
  • #18
62
0
Theonline program rejected it. Wasnt the other answer 389.39? because if it was, that was rejected as well
 
  • #19


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

Related Threads on Forces and Normal Force

  • Last Post
Replies
17
Views
5K
  • Last Post
Replies
7
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
3
Views
5K
Replies
1
Views
13K
  • Last Post
Replies
7
Views
867
  • Last Post
Replies
4
Views
3K
Top