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leianne Mar4-12 11:42 PM

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

Determine the magnitude of the horizontal force Fp required to push a block of a mass 30 kg, up along an inclined plane, with a constant acceleration of 0.8m/s2. The plane makes an angle of 25 to the horizontal and the coefficient of kinetic friction between the plane and the block is 0.21. (Hint: The solution involves setting up simultaneous equations in terms of x and y components of the unknown force, i.e. Fpcos25 and Fpsin25, and then solving them for Fp.)

2. Relevant equations

I asked my teacher how to do it but i didn't really understand him. He gave me this formula to find Fp. The answer given by the book is 250 N.

Fpcosθ = Fg +μ(mgcosθ + Fpsinθ) + Fa

where Fp= Force applied/of pull
Fa= Acceleration Force
Fg= Force due to gravity/weight

3. The attempt at a solution
I have plugged in the given but i still get the wrong answer (400+ N).

Can someone please explain this to me? Help me please. Please.

tonit Mar5-12 04:28 AM

Re: Friction help, please
 
First of all, you need to learn trigonometry to have these types of problems easier. I have attached an image where I have pointed out the most important parts of the problem.
At the top left corner you have the chosen directions for x and for y.

[itex]f[/itex] => force of friction (the red vector). The force of friction always is on the opposite side of the force being applied to the object.

[itex]G[/itex] => force of gravity (the orange vector). It is then divided into two components. The [itex]G[/itex][itex]x[/itex] component (the yellow vector) and [itex]G[/itex][itex]y[/itex] (the black vector). If you add those vectors you would get [itex]G[/itex] so it is fair to break it up in two components as it makes it easier to solve the problem. By trigonometry, I have written [itex]G[/itex][itex]x[/itex] = [itex]mgsin25[/itex] and [itex]G[/itex][itex]y[/itex] = [itex]mgcos25[/itex].

[itex]F[/itex]P => the force you apply (purple vector)

[itex]F[/itex]N => the normal force. (light green vector)

Now go on and try to solve it. Even though I got a result of [itex]200N[/itex], I believe it has been mistakenly put on the book [itex]250N[/itex].

tiny-tim Mar5-12 04:34 AM

welcome to pf!
 
hi leianne! welcome to pf! :smile:
Quote:

Quote by leianne (Post 3798924)
Fpcosθ = Fg +μ(mgcosθ + Fpsinθ) + Fa

where Fp= Force applied/of pull
Fa= Acceleration Force
Fg= Force due to gravity/weight

that Fg (which = mg) shouldn't be there
try again without it :smile:
(if you still don't understand why that is the formula, let us know)

leianne Mar5-12 05:06 AM

Re: Friction help, please
 
Quote:

Quote by tiny-tim (Post 3799188)
hi leianne! welcome to pf! :smile:


that Fg (which = mg) shouldn't be there
try again without it :smile:
(if you still don't understand why that is the formula, let us know)

Quote:

Quote by tonit (Post 3799181)
First of all, you need to learn trigonometry to have these types of problems easier. I have attached an image where I have pointed out the most important parts of the problem.
At the top left corner you have the chosen directions for x and for y.

[itex]f[/itex] => force of friction (the red vector). The force of friction always is on the opposite side of the force being applied to the object.

[itex]G[/itex] => force of gravity (the orange vector). It is then divided into two components. The [itex]G[/itex][itex]x[/itex] component (the yellow vector) and [itex]G[/itex][itex]y[/itex] (the black vector). If you add those vectors you would get [itex]G[/itex] so it is fair to break it up in two components as it makes it easier to solve the problem. By trigonometry, I have written [itex]G[/itex][itex]x[/itex] = [itex]mgsin25[/itex] and [itex]G[/itex][itex]y[/itex] = [itex]mgcos25[/itex].

[itex]F[/itex]P => the force you apply (purple vector)

[itex]F[/itex]N => the normal force. (light green vector)

Now go on and try to solve it. Even though I got a result of [itex]200N[/itex], I believe it has been mistakenly put on the book [itex]250N[/itex].


Thank you so much! If my memory doesn't fail me, I've actually gotten 200N and 250 N at one time. It's just that i didn't understand how i got it. :rofl:

Anyway i've plugged it in in the formula (without Fg) but my answer still comes up to 356.25 N :confused: Why is that? I don't get it im sorry. The formula i used is:

Fpcosθ = μ(mgcosθ + Fpsinθ) + Fa

Another question is that isn't friction acting in the opposite direction of the pull? Shouldn't that be negative then? Please help

leianne Mar5-12 05:24 AM

Re: Friction help, please
 
And if possible can you/anyone also explain to me how my teacher got that formula? Please bear with my little understanding. :( Thank you very much.

tonit Mar5-12 05:25 AM

Re: Friction help, please
 
did you see the image I've attached?

leianne Mar5-12 05:32 AM

Re: Friction help, please
 
Quote:

Quote by tonit (Post 3799230)
did you see the image I've attached?

Yes, I did. But isn't the question asking for the magnitude of the horizontal force? So shouldn't Fp also be broken down into components?

Anyway is your diagram showing that F = Fp - Gx - f? I've tried it and it gave me 204 N. yay am i annoying you? im really sorry.

tiny-tim Mar5-12 05:43 AM

(tonit, your diagram is clearly wrong, that is not being helpful :redface:)

hi leianne! :smile:
Quote:

Quote by leianne (Post 3799210)
i've plugged it in in the formula (without Fg) but my answer still comes up to 356.25 N :confused: Why is that? I don't get it im sorry. The formula i used is:

Fpcosθ = μ(mgcosθ + Fpsinθ) + Fa

can you show us exactly how you got from that formula to your answer? :smile:
Quote:

Quote by leianne (Post 3799229)
And if possible can you/anyone also explain to me how my teacher got that formula? Please bear with my little understanding. :( Thank you very much.

Quote:

Another question is that isn't friction acting in the opposite direction of the pull? Shouldn't that be negative then?
your teacher did F = ma along the slope

really, all the forces should be on the left, and only ma on the right

(in my opinion, it is really confusing to refer to "acceleration force" :frown:)

so, on the left, it is obvious that Fp is positive, but the other two are negative :smile:

leianne Mar5-12 05:58 AM

Re: Friction help, please
 
Quote:

Quote by tiny-tim (Post 3799244)
(tonit, your diagram is clearly wrong, that is not being helpful :redface:)

hi leianne! :smile:


can you show us exactly how you got from that formula to your answer? :smile:



your teacher did F = ma along the slope

really, all the forces should be on the left, and only ma on the right

(in my opinion, it is really confusing to refer to "acceleration force" :frown:)

so, on the left, it is obvious that Fp is positive, but the other two are negative :smile:


Fpcosθ = μ(mgcosθ + Fpsinθ) + Fa
Fpcos(25) = 0.21[(30x9.81xcos25) + Fpsin(25)] + 30x0.8
0.9Fp = 0.21(266.7 + 0.4Fp) + 24
0.9Fp = 56 + 0.084Fp + 24
0.9Fp - 0.084Fp = 56 + 24
0.816Fp/0.816 = 80/0.816
Fp = 98 N (what?!:eek:)

yay i think i got 356 N from inventing/mixing up formulas. wah what should i do?

anyway with what you last said, will the formula then be :

Fp - Fgx - f = ma?

tiny-tim Mar5-12 06:12 AM

oops! :redface:

i didn't draw a diagram, and i left out an mgsinθ (the x component of the weight)

try that :smile:

leianne Mar5-12 06:16 AM

Re: Friction help, please
 
Quote:

Quote by tiny-tim (Post 3799266)
oops! :redface:

i didn't draw a diagram, and i left out an mgsinθ (the x component of the weight)

try that :smile:


Well actually the Fgx in my Fp - Fgx - f = ma is equal to mgsinθ. I just called it Fgx since it's the horizontal component of the weight. And the answer i kept getting using that formula is 204 N which means im still missing 46 N (according from the answer at the back of my book. oh yes, my book is Engineering Mechanics by val ivanoff)

tiny-tim Mar5-12 06:25 AM

Quote:

Quote by leianne (Post 3799272)
And the answer i kept getting using that formula is 204 N

no, that should be 204/.816, shouldn't it? :wink:

leianne Mar5-12 06:29 AM

Re: Friction help, please
 
Quote:

Quote by tiny-tim (Post 3799280)
no, that should be 204/.816, shouldn't it? :wink:

WOAH it's 250 N!!!!!! finally!!!! that is awesome yay thank you so much!!! but what is the right formula then? how should i show that? again thank you very very much yay!!!

is it simply :
ma = Fp - mgsin(theta) - f
ma = (Fpcostheta - Fpsintheta) - mgsintheta - f ?

leianne Mar5-12 06:36 AM

Re: Friction help, please
 
Quote:

Quote by tiny-tim (Post 3799280)
no, that should be 204/.816, shouldn't it? :wink:

yay i get it now!!! thank you very very very much!!! yay im just genuinely ecstatic right now and really really grateful!!! thank you thank you thank you!!!

tiny-tim Mar5-12 06:37 AM

Quote:

Quote by leianne (Post 3799287)
but what is the right formula then? how should i show that?

Pcosθ - mgsinθ - μ(mgcosθ + Psinθ) = ma :smile:

(ie, all the x-components on the LHS, and ma on the RHS)

leianne Mar5-12 06:46 AM

Re: Friction help, please
 
Quote:

Quote by tiny-tim (Post 3799298)
Pcosθ - mgsinθ - μ(mgcosθ + Psinθ) = ma :smile:

(ie, all the x-components on the LHS, and ma on the RHS)

yes i get it now yay and it's all thanks to you!!!! thank you very very very much!!!! i feel kind of embarrassed now that i've figured out what my silly mistake was :redface: i understand everything now and i better burn it to my memory yay!!! really thank you very very very much!!!


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