
#1
Mar412, 11:42 PM

P: 9

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/s^{2}. 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_{p}cos25° and F_{p}sin25°, 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. F_{p}cosθ = F_{g} +μ(mgcosθ + F_{p}sinθ) + F_{a} 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. 



#2
Mar512, 04:28 AM

P: 55

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]. 



#3
Mar512, 04:34 AM

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P: 26,167

hi leianne! welcome to pf!
try again without it(if you still don't understand why that is the formula, let us know) 



#4
Mar512, 05:06 AM

P: 9

Friction help, pleaseThank 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: 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
Mar512, 05:24 AM

P: 9

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
Mar512, 05:25 AM

P: 55

did you see the image I've attached?




#7
Mar512, 05:32 AM

P: 9

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
Mar512, 05:43 AM

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P: 26,167

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



#9
Mar512, 05:58 AM

P: 9

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
Mar512, 06:12 AM

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P: 26,167

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



#11
Mar512, 06:16 AM

P: 9

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
Mar512, 06:25 AM

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P: 26,167





#13
Mar512, 06:29 AM

P: 9

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



#14
Mar512, 06:36 AM

P: 9





#15
Mar512, 06:37 AM

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P: 26,167

(ie, all the xcomponents on the LHS, and ma on the RHS) 



#16
Mar512, 06:46 AM

P: 9




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