Strange physics question. Help

  • #1

Homework Statement


Two boxes each mass 10 kg are raised 2.0m to a shelf. The first one is lifted and second one
is pushed up a smooth ramp(frictionless). If the applied force on the second box is 50N, calculator the angle between the ramp
and the ground

Homework Equations


n/a


The Attempt at a Solution


Got work of the first box
(10)(9.8)(2)=W
=196J
Don't know what to do with it, tried 198=50(2)sintheta, but that won't work. Please show full solution, lost.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31

The Attempt at a Solution


Got work of the first box
(10)(9.8)(2)=W
=196J
Don't know what to do with it, tried 198=50(2)sintheta, but that won't work. Please show full solution, lost.

You'd need to account for the component of the 10 kg weight opposing the 50 N force.


Remember that sine = opposite/hypotenuse

so the distance the mass is sliding up the ramp is the hypotenuse.
 
  • #3
11,872
5,526
[strike]You can determine the force to raise the first box via F=mg right?

The same amount of vertical force is needed to raise the second box on the ramp.

So try drawing a force diagram of a force pushing/pulling a block up a ramp.[/strike]

EDIT: disregard this post... see followon post on ramp mechanical advantage
 
Last edited:
  • #4
You can determine the force to raise the first box via F=mg right?

The same amount of vertical force is needed to raise the second box on the ramp.

So try drawing a force diagram of a force pushing/pulling a block up a ramp.
so the force requried to life the first box is 98N,
so 98=sin(90)
=98

what about the other one, it wouldn't be 50=sintheta right?
 
  • #6
Try reading about mechanical advantage from ramps in this article:

http://www.teachengineering.org/vie..._simp_machines/cub_simp_machines_lesson01.xml

I think now you were on the right track with the work calculation for lifting and now with the ramp's assistance you can use less force to do the same work.

Here's a freebody diagram that may help too:

http://en.wikipedia.org/wiki/File:Free_body.svg
Yea I looked at it and do I'm still lost, could you please tell me how to find it
 
  • #7
11,872
5,526
Yea I looked at it and do I'm still lost, could you please tell me how to find it
We're here to help you when you get stuck not to do your homework. You need to show more work before we can help. I think what I gave would be sufficient to solve the problem.
 
  • #8
We're here to help you when you get stuck not to do your homework. You need to show more work before we can help. I think what I gave would be sufficient to solve the problem.
okay sorry, I figured out 196J=fxd
196=50*d
3.92=d
I now have the opposite and the hyp. I did sin^-1(2/3.92)=30degrees
I double checked it with 50*3.92*cos30degrees= 168J
This is not the 196 joules needed, where did I go wrong?
 
  • #9
11,872
5,526
okay sorry, I figured out 196J=fxd
196=50*d
3.92=d
I now have the opposite and the hyp. I did sin^-1(2/3.92)=30degrees
I double checked it with 50*3.92*cos30degrees= 168J
This is not the 196 joules needed, where did I go wrong?
I think your check is wrong. How did you come up with it?
 
  • #10
I think your check is wrong. How did you come up with it?
Since I know the formula for work W=Fxdcostheta, I wanted to make sure the that about of work would be the same as for the case of the ramp and the case of the box being lifted up 90degrees. I figured that the side length must be 3.92 and used that as they hyp. I used sin to figure out the angle then checked it with that angle, but the work was 168 instead of the needed 196. I don't know what I did wrong. could you point it out?
 
  • #11
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
Since I know the formula for work W=Fxdcostheta
But it isn't.

The "formula" is always W=Fd, where F is the force and d is the distance moved in the direction of that force. The formula never varies from this.

You have a force acting along the slope here. What is its magnitude, and through what distance does the point of application of that force move?

Your only "mistake" is to have become too complacent in your use of that "formula". :wink:

Stay alert, or you'll be tripped up. Again, and when you least expect it. :smile:
 
  • #12
223
10
Sketch out the information you know with pen and paper. That way any assignment will become much less difficult to tackle.
Calculate the hypotenuse, which is the distance the 2nd box is being pushed up to reach the shelf.

On your sketch, mark every force you know of, that is acting on the 2nd box. There is gravity, the force it is being pushed by and the force the surface of the ramp is exerting on the box and luckily you won't have to account for friction.
If you think about it, the surface of the ramp does work as well all the way the box is being pushed. A = Fs or W=Fd if you prefer. What is the force the ramp is exerting on the box?

Also, do you understand why you are equating the potential energy of the 1st box already on the shelf to the work done by the force pushing the 2nd box to reach the shelf?
 
  • #13
139
7

Homework Statement


Two boxes each mass 10 kg are raised 2.0m to a shelf. The first one is lifted and second one
is pushed up a smooth ramp(frictionless). If the applied force on the second box is 50N, calculator the angle between the ramp
and the ground

Homework Equations


n/a


The Attempt at a Solution


Got work of the first box
(10)(9.8)(2)=W
=196J
Don't know what to do with it, tried 198=50(2)sintheta, but that won't work. Please show full solution, lost.
The work that you have to do is

[itex]\textit{W}[/itex]= [itex]\textit{m}[/itex][itex]\cdot[/itex][itex]\textit{g}[/itex][itex]\cdot[/itex][itex]\textit{h}[/itex] = [itex]\textit{50}[/itex][itex]\cdot\frac{\textit{h}}{\textit{sin Θ}}[/itex]
 

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