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Constant velocity and newton

  • Thread starter flanders
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  • #1
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Homework Statement


Hi. My task is to find the Force F, so that velocity is constant.

I know that the velocity is constant, therefore the sum of all forces must be zero!

On the image I have used vector decomposition (the black lines). The orange ones are the vectors.

I know that the angle is 36,9 degrees. (the one marked with a half a circle).

I also know that the friction number is 0,25.


Explanation of the attached image:

G = Gravity

N = Normal Force

Ff = Friction

F = The force I must find





Homework Equations




[tex]Gx=0,6mg[/tex]
[tex]Gy=0,8mg[/tex]

[tex]Fx=0,8ma[/tex]
[tex]Fy=0,6ma[/tex]

[tex]\mu \; =\; 0.25[/tex]

[tex]N = Gy + Fy[/tex]


The Attempt at a Solution



I have tried to find N by using this formula:

[tex]\mu \; =\; \frac{Ff}{N}[/tex]

Then I thought that I could find Fy and Fx and use Phytagoras to find F. But my idea doesn't work (at least not the way I did it).

I don't know how to think to solve this one... The mass is unknown and frustrates me alot. Can anyone give me a tip?
 

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Answers and Replies

  • #2
tiny-tim
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hi flanders! :wink:

have you tried drawing a vector triangle ?

but if you don't know how to do that, then start the equations with F = ma in the normal direction …

you know that a = 0, so N = … ? :smile:

(and call the mass "m" … it'll cancel out in the end anyway)
 
  • #3
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I got curious in how you do the vector triangle? You just put the vectors one after another and then you'll find out that you get back to the starting point - therefore the velocity must be constant? And what will the different angles look like? Hm... I'm sorry, but I don't understand that much...
 
  • #4
tiny-tim
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I got curious in how you do the vector triangle? You just put the vectors one after another and then you'll find out that you get back to the starting point -
yes, you have (in this case) a right-angled triangle, with sides N mg and (F minus friction) …

it works because you know the acceleration is zero, so the forces must add to zero, and forces add like vectors (head-to-tail), so the triangle must be closed

since you know two angles and one side (mg), you can immediately find the other two sides … then use friction = µN to subtract the friction from (F minus friciton) to get F :smile:
 
  • #5
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Great!! Thank you - I think I understood a little bit more! :)

Now I used tan to find F, and then Sin to find N. As a result I got F = 4,30m.

Is it possible to do something with the m..? I don't think we are able to find it with that sort of information we've got?
 
  • #6
tiny-tim
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hi flanders! :smile:

(just got up :zzz: …)
Now I used tan to find F, and then Sin to find N. As a result I got F = 4,30m.
i don't understand that :confused:
Is it possible to do something with the m..? I don't think we are able to find it with that sort of information we've got?
no, you're right …

the question asks for the applied force F, and for that we do need to know m :redface:
 
  • #7
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Adding all the "main" vectors gives us a triangle with angles of 90, 36.9 and 53.1 degrees.

As seen on the drawed picture attached, I have:

[TEX]N\; =\; \frac{G}{\mbox{C}os\left( 36.9 \right)}\; =\; 1.25G[/TEX]

[tex]F\; =\; N\cdot \sin \; \left( 36.9 \right)\; =\; 1.25G\cdot \sin \; \left( 36.9 \right)\; =\; 0.75G[/tex]

Then I find F(Friction):

Ff = 0.25*1.25G = 0.313G

F-Ffriction will then be:

[tex]F\; =\; F-Ffriction\; =\; 0.75G\; -\; 0.313G\; =\; 0.46G[/tex]

/Flanders
 

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  • #8
tiny-tim
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ah, but isn't F supposed to be along the slope?

if so, the triangle should show F at an angle, with the right-angle at top-left instead of top-right :wink:
 
  • #9
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I'm sorry - I didn't specify the direction of the force, but we're supposed to find the horizontal force.

Thank you very much, it was helpfull :)
 
  • #10
tiny-tim
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I'm sorry - I didn't specify the direction of the force, but we're supposed to find the horizontal force.
hold it!

i don't think the vector triangle method works for forces in four different directions …

(i assumed there were only three different directions, so we could lump F and friction together, to make a triangle, and then subtract the friction)

you can draw a vector quadrilateral, and it will work (you'll have to draw the friction equal to µ times N), or it might be just as easy to use x and y components
 
  • #11
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Hey Tiny-Tim!

I've tried to set up two equations (X and Y-direction) and to solve it without results :(.

Could you please explain the quadrilateral-method in short terms? I would really appreciate it =).

It should also be said, that the answer is to be given in m*g. For instance 0.8mg.

The force is still horizontal.

Update: I got an answer now... Tried to set up the equations one more time and found that F = 0.8G/0.65, which is 1.23G. It sounds right, doesn't it?
 
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  • #12
SammyS
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Homework Statement


Hi. My task is to find the Force F, so that velocity is constant.

I know that the velocity is constant, therefore the sum of all forces must be zero!

On the image I have used vector decomposition (the black lines). The orange ones are the vectors.

I know that the angle is 36,9 degrees. (the one marked with a half a circle).

I also know that the friction number is 0,25.

Explanation of the attached image:

G = Gravity

N = Normal Force

Ff = Friction

F = The force I must find

Homework Equations



[tex]Gx=0,6mg[/tex]
[tex]Gy=0,8mg[/tex]

[tex]Fx=0,8ma[/tex]
[tex]Fy=0,6ma[/tex]

[tex]\mu \; =\; 0.25[/tex]

[tex]N = Gy + Fy[/tex]

The Attempt at a Solution



I have tried to find N by using this formula:

[tex]\mu \; =\; \frac{Ff}{N}[/tex]

Then I thought that I could find Fy and Fx and use Pythagoras to find F. But my idea doesn't work (at least not the way I did it).

I don't know how to think to solve this one... The mass is unknown and frustrates me alot. Can anyone give me a tip?
The following two equations can't be correct. If a=0, then F=0.
[tex]F_x=0,8ma[/tex]
[tex]F_y=0,6ma[/tex]

They should be:
[tex]F_x=0,8|\vec{F}|[/tex]
[tex]F_y=-0,6|\vec{F}|[/tex]

Be careful of signs: I would write: Gx = ‒0,6mg and Gy = ‒0,8mg

Then the y-components give N = ‒Gy‒Fy → N = 0,8mg + 0,6|F| .

The x-components give: Fx+Gx+Ff=0 → 0,8|F|‒0,6mg‒μN=0.

Two equations in three unknowns.

Find F in terms of mass m.
 
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  • #13
tiny-tim
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Could you please explain the quadrilateral-method in short terms? I would really appreciate it =).
I think we'd better make sure you can do it the usual way first (I'll explain the quadrilateral later) …
I've tried to set up two equations (X and Y-direction) and to solve it without results :(.
Show us your equations. :smile:
 
  • #14
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SammyS: I have the same results as you have =). I have written N=Gy+Fy, so that they will have their "correct" direction... I think!

tiny-tim: I have done it the same way as SammyS:
Then the y-components give N = ‒Gy‒Fy → N = 0,8mg + 0,6|F| .

The x-components give: Fx+Gx+Ff=0 → 0,8|F|‒0,6mg‒μN=0.
I did then put N into the x-equation:

0.8F-0.6mg-0.25(0.8mg+0.6F)=0

calculate and it would be:

0.8F-0.6mg-0.2mg-0.15F=0

F = (0.8mg)/0.65

Does it look right to you?

Thanks for the help so far! I'm (hopefully!!) starting to see things clearer!
 
  • #15
tiny-tim
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Yes, that looks good! :smile:

Now try drawing a quadrilateral with those magnitudes, and see if it joins up. :wink:

(that's a good way of checking the answer!)
 

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