How Long Should the Inclined Plane Be for a 100 kg Box?

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Homework Help Overview

The discussion revolves around determining the length of an inclined plane required to raise a 100 kg box to a height of 5 m using a force of 250 N. Participants explore the relationship between force, weight, height, and the length of the incline, while grappling with the absence of a specified angle of inclination.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need to incorporate trigonometric functions such as sinθ or cosθ due to the lack of an angle. There are attempts to derive relationships between the forces acting on the box and the incline's geometry. Some participants express confusion about how to proceed without a defined angle, while others suggest drawing diagrams to visualize the forces involved.

Discussion Status

The conversation is ongoing, with participants providing guidance on drawing free body diagrams and considering the forces acting on the box. There is exploration of various interpretations regarding the incline's angle and its implications on the calculations. Some participants have shared equations and attempted calculations, while others are still seeking clarity on certain aspects.

Contextual Notes

Participants note that the problem does not provide an angle for the incline, which complicates the calculations. There is mention of constraints regarding the maximum angle that can be used without losing the ability to accelerate the mass up the incline.

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Homework Statement


  1. An inclined plane is used to raise a 100 kg box to a height of 5 m. How long must the inclined plane be if a force of 250 N is used to push the box up the inclined plane?

Homework Equations


i have search and searched all over google to try to figure this out.. I am not sure if I'm over thinking it or just confused myself even more. I found one equation that I think is correct, but as i was looking at other examples I tried to use this formula and it never worked out like I thought it should. Also, I am always finding examples that have an angle of inclination.. But the formula I think should work is.

\frac{f}{w}=\frac{h}{l}

The Attempt at a Solution


\frac{250}{100}=\frac{5}{l}
l=2?

cheers
keith
 
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Hi keithcuda. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

You won't be able to avoid involving either sinθ or cosθ in your calculations.
Can you attach a pic showing the relevant diagram you have drawn for this problem?
 
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You could draw a free body diagram to see what forces are opposing the motion of the box on an incline. For there to be an upward motion, that or those force(s) must be less than the upward force.
 
i can try to draw pic and post it...

That is where i am confused.. the question isn't giving me an angle of the incline.. so how do i include sinθ or cosθ?
 
Maybe one of the opposing forces is related to the angle ##\theta## of the incline.
 
θ
keithcuda said:
i can try to draw pic and post it...

That is where i am confused.. the question isn't giving me an angle of the incline
So call the slope θ, or x or something.
 
NascentOxygen said:
Hi keithcuda. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

You won't be able to avoid involving either sinθ or cosθ in your calculations.
Can you attach a pic showing the relevant diagram you have drawn for this problem?
?#3.jpg
 

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OK. Next show a vector representing the force of gravity on that box. What magnitude would this force have? And its units?

Have you given the slope a name, i.e., a letter?
 
OK, so
NascentOxygen said:
OK. Next show a vector representing the force of gravity on that box. What magnitude would this force have? And its units?

Have you given the slope a name, i.e., a letter?

Ok, so I did some more googling...
I found this, Fp=w(h/l)
Fp=250N
w=mg=100*9.8
h=5
l=slope
250=980(5/l)
l=19.6m ?
 
  • #10
let me know what I am doing wrong or missing and as soon as i leave work and get home I'll work on it some more

cheers
 
  • #11
Once you have answered the questions in post #8 ...

gravitational force on the box can be resolved into an equivalent pair of forces at right-angles to each other: one force acting parallel to the plane, and the second force acting perpendicular to the plane

Forces are vectors, so to add them you must add them as vectors. Draw this "triangle of forces" to show these two perpendicular forces summing to equal the gravitational force on the box. Use trigonometry to calculate the magnitudes of the unknown sides of this triangle.
 
  • #12
Alright, I believe I know what you are talking about... But, I'm just getting home from my power plant after working a 17 hour shift. Once I get some sleep, I'll work on post #8 and #11.
 
  • #13
If you need help with the vector diagram, search on google images for "block on inclined plane". You'll get hundreds. :w

http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg
 
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  • #14
NascentOxygen said:
If you need help with the vector diagram, search on google images for "block on inclined plane". You'll get hundreds. :w

http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg
Thank you, I'm actually working on it right now :)
 
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  • #15
OK, so here is what I put together so far...
incline_pdf.jpg
 
  • #16
So far so good ...
 
  • #17
NascentOxygen said:
So far so good ...
I guess this is where I am really stuck. Without a given angel for theta, how to I proceed with my problem?
 
  • #18
g assumed at 9.81 m/s/s
Choose any incline angle (and therefore incline length) but don't exceed 14.764 ° incline though, as you won't have any force left to accelerate the mass up the incline.
The problem is, if the force is kept constant, at the top of the slope, the mass will still be moving (and accelerating).

So, you accelerate the mass so far up the slope then remove the (250 N) force and allow it to decelerate under gravity, to come to a rest at the top of the slope.
Ive ran a few examples through, and found that :
Regardless of the incline length, the length you apply the force over turns out to be the same.
The key is that you put the work in ( force * distance) which is equal to the final PE (m*g*h)
So :
250 * d = m * g * h
You may do the rest.
 
  • #19
keithcuda said:
I guess this is where I am really stuck. Without a given angel for theta, how to I proceed with my problem?
The problem includes a force 250N. Of the expressions you have derived, what can you equate to 250N?
 
  • #20
dean barry said:
g assumed at 9.81 m/s/s
Choose any incline angle (and therefore incline length) but don't exceed 14.764 ° incline though, as you won't have any force left to accelerate the mass up the incline.
The problem is, if the force is kept constant, at the top of the slope, the mass will still be moving (and accelerating).

So, you accelerate the mass so far up the slope then remove the (250 N) force and allow it to decelerate under gravity, to come to a rest at the top of the slope.
Ive ran a few examples through, and found that :
Regardless of the incline length, the length you apply the force over turns out to be the same.
The key is that you put the work in ( force * distance) which is equal to the final PE (m*g*h)
So :
250 * d = m * g * h
You may do the rest.
so, 250*d=100*9.8*5
d=4900/250
d=19.6
 
  • #21
NascentOxygen said:
The problem includes a force 250N. Of the expressions you have derived, what can you equate to 250N?
I am sorry, but I'm not sure I understand what you are asking :(
 
  • #22
keithcuda said:
I am sorry, but I'm not sure I understand what you are asking :(
On your figure you show the pink 250N acting up the slope. You equate this to the blue component of weight acting in the exact opposite direction. You have the equation for that, you called it fg∥.
 
  • #23
Thats right, what you find is that if you remove the force at that point, it will decelerate to a stop exactly at the top of the slope.
This d applies regardless of which incline angle you choose ( up to 14.764 °).

Dean Barry
 
  • #24
NascentOxygen said:
On your figure you show the pink 250N acting up the slope. You equate this to the blue component of weight acting in the exact opposite direction. You have the equation for that, you called it fg∥.
ok so, 250=980(sinθ)
sin(θ)=250/980
θ=sin-1(250/980)
θ=14.77
 
  • #25
so if θ=14.77°
then, sin(14.8)=5/h
h=5/sin(14.8)
h=19.57m
 
  • #26
Looks right. Note: the length of the slope is not the "l" you show on your figure.

In 99% of the problems of this type you are required to find the slope or θ. In today's problem you don't specifically have to. Going back to your figure, you could have looked at similar triangles: the slope triangle, and the weight triangle. This leads to the equating of ratios (which may be what you copied from something on the net way back near the start).

It would be worthwhile making sure you can see those similar triangles, before leaving this problem.
 
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  • #27
keithcuda said:
I guess this is where I am really stuck. Without a given angel[/color] for ...
Fingers crossed that you'll have that angel on your shoulder during exams! o_O

http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg
 
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  • #28
NascentOxygen said:
Fingers crossed that you'll have that angel on your shoulder during exams! o_O

http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg
haha, good catch ;)

I tend to type too fast and not proofread enough...
Thank you for your help. I appreciate it!
Happy Holidays
Cheers
 
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  • #29
dean barry said:
The key is that you put the work in ( force * distance) which is equal to the final PE (m*g*h)
So :
250 * d = m * g * h
You may do the rest.
That's an interesting alternative way to work this problem. Perfectly valid, and a good way to confirm ones understanding of kinetics. Well done.
 
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  • #30
Heres the excel check sheet (tidied up version) for proof.
Note, excel calculates in radians.
The tan cells are inputs (vary if you wish).
Dont exceed the maximum slope angle (Am)
(ive used 10° as an example , alter as you wish)
The two blue cells match regardless of the slope value.
Regards.
Dean
 

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