Friction help, please

leianne

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. F_pcos25° and F_psin25°, 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.


where F_p= Force applied/of pull
F_a= Acceleration Force
F_g= 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

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

hi leianne! welcome to pf!
that F_g (which = mg) shouldn't be there

try again without it

(if you still don't understand why that is the formula, let us know)

leianne

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.

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


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

leianne

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

did you see the image I've attached?

leianne

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

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

hi leianne!
can you show us exactly how you got from that formula to your answer?
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" )

so, on the left, it is obvious that F_p is positive, but the other two are negative

leianne

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?!)

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 :

tiny-tim

oops!

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

try that

leianne

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

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

leianne

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

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

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

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

leianne

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 i understand everything now and i better burn it to my memory yay!!! really thank you very very very much!!!