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

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In summary, the student attempted to solve for the magnitude of the horizontal force 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.
  • #1
leianne
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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. Fpcos25° and Fpsin25°, 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.

Fpcosθ = Fg +μ(mgcosθ + Fpsinθ) + Fa

where Fp= Force applied/of pull
Fa= Acceleration Force
Fg= 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.
 
Last edited:
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  • #2
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].
 
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  • #3
welcome to pf!

hi leianne! welcome to pf! :smile:
leianne said:
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)
 
  • #4
tiny-tim said:
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)

tonit said:
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 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
 
  • #5
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
did you see the image I've attached?
 
  • #7
tonit said:
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.
 
  • #8
(tonit, your diagram is clearly wrong, that is not being helpful :redface:)

hi leianne! :smile:
leianne said:
… 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 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? :smile:
leianne said:
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" :frown:)

so, on the left, it is obvious that Fp is positive, but the other two are negative :smile:
 
  • #9
tiny-tim said:
(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?
 
  • #10
oops! :redface:

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

try that :smile:
 
  • #11
tiny-tim said:
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 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)
 
  • #12
leianne said:
And the answer i kept getting using that formula is 204 N

no, that should be 204/.816, shouldn't it? :wink:
 
  • #13
tiny-tim said:
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 ?
 
  • #14
tiny-tim said:
no, that should be 204/.816, shouldn't it? :wink:

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!
 
  • #15
leianne said:
… 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)
 
  • #16
tiny-tim said:
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!
 

What is friction?

Friction is a force that resists the relative motion or tendency of motion between two surfaces in contact.

Why is friction important?

Friction plays a crucial role in our daily lives, as it allows us to walk, drive, and use tools without slipping. It also helps to slow down or stop moving objects, which is essential for safety.

What factors affect friction?

The main factors that affect friction are the roughness of the surfaces in contact, the force pressing the surfaces together, and the type of material the surfaces are made of.

How can friction be reduced?

Friction can be reduced by using lubricants, such as oil or grease, between the surfaces in contact. Smoother surfaces and reducing the force pressing the surfaces together can also decrease friction.

What are the different types of friction?

There are three types of friction: static, kinetic, and rolling. Static friction occurs when two surfaces are not moving relative to each other, kinetic friction occurs when two surfaces are moving relative to each other, and rolling friction occurs when one surface is rolling over another surface.

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