Calculating Fp for a Block on an Inclined Plane | Friction Help"

[leianne]

Homework Statement

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.)

Homework 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_pcosθ = F_g +μ(mgcosθ + F_psinθ) + F_a​

where F_p= Force applied/of pull
F_a= Acceleration Force
F_g= Force due to gravity/weight

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.

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.

 

[tiny-tim]

welcome to pf!

hi leianne! welcome to pf!

F_pcosθ = F_g +μ(mgcosθ + F_psinθ) + F_a​

where F_p= Force applied/of pull
F_a= Acceleration Force
F_g= Force due to gravity/weight

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]

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)

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.

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 I am 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]

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]

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? I am really sorry.

 

[tiny-tim]

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

hi leianne!

… 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 I am 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?

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.

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" )

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

 

[leianne]

(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

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? ​

 

[tiny-tim]

oops!

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

try that

 

[leianne]

oops!

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

try that

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 I am 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]

And the answer i kept getting using that formula is 204 N

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

 

[leianne]

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

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]

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

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

 

[tiny-tim]

… but what is the right formula then? how should i show that?

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

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

 

[leianne]

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

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