How can I find the force needed for constant velocity with known variables?

AI Thread Summary
To find the force needed for constant velocity, the sum of all forces must equal zero, indicating that friction and the applied force must balance out. The discussion involves using vector decomposition and equations related to gravitational and frictional forces, with a known friction coefficient of 0.25 and an angle of 36.9 degrees. Participants suggest using vector triangles and quadrilaterals to visualize the forces and derive equations for the normal force and applied force. Ultimately, the calculated force is expressed in terms of gravitational force, leading to a final result of F = (0.8mg)/0.65. The conversation emphasizes the importance of understanding vector relationships in solving the problem.
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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




Gx=0,6mg
Gy=0,8mg

Fx=0,8ma
Fy=0,6ma

\mu \; =\; 0.25

N = Gy + Fy


The Attempt at a Solution



I have tried to find N by using this formula:

\mu \; =\; \frac{Ff}{N}

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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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)
 
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...
 
flanders said:
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:
 
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?
 
hi flanders! :smile:

(just got up :zzz: …)
flanders said:
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:
 
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:

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

F\; =\; N\cdot \sin \; \left( 36.9 \right)\; =\; 1.25G\cdot \sin \; \left( 36.9 \right)\; =\; 0.75G

Then I find F(Friction):

Ff = 0.25*1.25G = 0.313G

F-Ffriction will then be:

F\; =\; F-Ffriction\; =\; 0.75G\; -\; 0.313G\; =\; 0.46G

/Flanders
 

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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:
 
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
flanders said:
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
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
flanders said:

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



Gx=0,6mg
Gy=0,8mg

Fx=0,8ma
Fy=0,6ma

\mu \; =\; 0.25

N = Gy + Fy

The Attempt at a Solution



I have tried to find N by using this formula:

\mu \; =\; \frac{Ff}{N}

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.
F_x=0,8ma
F_y=0,6ma

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

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
flanders said:
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
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
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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