1. Mar 4, 2012

### 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. 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).

Last edited: Mar 4, 2012
2. Mar 5, 2012

### 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.

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

$G$ => force of gravity (the orange vector). It is then divided into two components. The $G$$x$ component (the yellow vector) and $G$$y$ (the black vector). If you add those vectors you would get $G$ 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 $G$$x$ = $mgsin25$ and $G$$y$ = $mgcos25$.

$F$P => the force you apply (purple vector)

$F$N => the normal force. (light green vector)

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

Last edited: Jun 16, 2012
3. Mar 5, 2012

### tiny-tim

welcome to pf!

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

try again without it

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

4. Mar 5, 2012

### 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. :rofl:

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:

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

5. Mar 5, 2012

### 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.

6. Mar 5, 2012

### tonit

did you see the image I've attached?

7. Mar 5, 2012

### 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.

8. Mar 5, 2012

### 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 Fp is positive, but the other two are negative

9. Mar 5, 2012

### 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 :

Fp - Fgx - f = ma?

10. Mar 5, 2012

### tiny-tim

oops!

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

try that

11. Mar 5, 2012

### 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)

12. Mar 5, 2012

### tiny-tim

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

13. Mar 5, 2012

### 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 ?

14. Mar 5, 2012

### 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!!!

15. Mar 5, 2012

### tiny-tim

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

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

16. Mar 5, 2012

### 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!!!