Calculating Minimum Force for Turning a Barrel over a Step

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Homework Help Overview

The problem involves calculating the minimum horizontal force required to push a barrel over a step. The context includes principles of moments and torque, as well as geometric considerations related to the barrel's dimensions and the step's height.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the principle of moments and torque calculations, questioning the height at which the force is applied and the location of the pivot point. There are attempts to clarify the geometry involved, including the distances relevant to the barrel and the step.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem setup. Some have offered insights into the geometric relationships and the need for trigonometric considerations, while others express uncertainty about the calculations and seek alternative methods.

Contextual Notes

There are mentions of potential confusion regarding the dimensions of the barrel and the step, as well as the application of trigonometry versus geometry in solving the problem. Participants note that they have not all learned trigonometry, which may affect their ability to engage with certain suggested methods.

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Homework Statement



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Find the smallest horizontal force required to push the barrel over the step.



Homework Equations



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The Attempt at a Solution



By the principle of moments,
F x 0.2 = 1500 x 0.5 .
Thus, F = 3750N.

I think that there's a smaller force, right??
 
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tgpnlyt7095 said:
By the principle of moments,
F x 0.2 = 1500 x 0.5 .
What's the height of the force F? Where's your pivot?
 
Doc Al said:
What's the height of the force F? Where's your pivot?

the height is 0.2 + 0.5 = 0.7m.

The pivot , i suppose is the edge of the step ??
 
tgpnlyt7095 said:
the height is 0.2 + 0.5 = 0.7m.
Your diagram shows the force level with the center of the barrel. I guess that's wrong? And the height of the step is labeled as 0.2, but it's drawn as being at the height of the force.

What's the height of the step? Where is the force applied? At h = .5? or .7?

The pivot , i suppose is the edge of the step ??
Right. So what's the moment arm (distance) for computing the torque due to F?
 
Doc Al said:
Your diagram shows the force level with the center of the barrel. I guess that's wrong? And the height of the step is labeled as 0.2, but it's drawn as being at the height of the force.

What's the height of the step? Where is the force applied? At h = .5? or .7?Right. So what's the moment arm (distance) for computing the torque due to F?

My apologies. the height of the step is 0.2m . And 0.5 represents the radius of the circle.
The force is applied at the height of 0.7m
 
tgpnlyt7095 said:
My apologies. the height of the step is 0.2m . And 0.5 represents the radius of the circle.
The force is applied at the height of 0.7m
Excellent. So what's the torque due to that force about the pivot? And what's the torque due to the weight of the barrel?
 
Doc Al said:
Excellent. So what's the torque due to that force about the pivot?

It should be, F x D = 1500N x 0.7m = 1050Nm
 
  • #10
tgpnlyt7095 said:
It should be, F x D = 1500N x 0.7m = 1050Nm
What's the value of D? I assume the barrel weighs 1500 N? How far is the center of the barrel from the edge of the step?
 
  • #11
Doc Al said:
What's the value of D? I assume the barrel weighs 1500 N? How far is the center of the barrel from the edge of the step?

Distance of Barrel from the centre to the edge of the step
= 0.5m

so is it F x D = 1500N x 0.5m
= 750Nm ??
 
  • #12
the weight (1500N) is not perpendicular to the distance so it's not 1500N x 0.5m.

you need to use some trigo here, i think (unless there's a shortcut method).
 
  • #13
arkofnoah said:
the weight (1500N) is not perpendicular to the distance so you can't do 1500N x 0.5m

But i thought its a circle ??
 
  • #14
I think i got it.

F x D , = 1500N x ( 0.5m - 0.2m )
= 450Nm
 
  • #15
tgpnlyt7095 said:
Distance of Barrel from the centre to the edge of the step
= 0.5m
No. That's the radius of the barrel. (Use a bit of trig/geometry.)

so is it F x D = 1500N x 0.5m
= 750Nm ??
Again, D is the distance between the line of the applied force F and the pivot. So what is D?
 
  • #16
Doc Al said:
No. That's the radius of the barrel. (Use a bit of trig/geometry.)


Again, D is the distance between the line of the applied force F and the pivot. So what is D?

0.5m - 0.2m = 0.3m
 
  • #17
Doc Al said:
No. That's the radius of the barrel. (Use a bit of trig/geometry.)

actually the distance of Barrel from the centre to the edge of the step is the radius of the circle :-p
 
  • #18
arkofnoah said:
actually the distance of Barrel from the centre to the edge of the step is the radius of the circle :-p
LOL... yeah, I should have said "What is the horizontal distance from barrel center to the edge of the step". :redface:
 
  • #19
Sorry .. ;(
 
  • #20
tgpnlyt7095 said:
0.5m - 0.2m = 0.3m
What's the height of F? What's the height of the step? The vertical distance between them is D.
 
  • #21
Doc Al said:
What's the height of F? What's the height of the step? The vertical distance between them is D.

Okay, so the torque is 1500 x 0.3 = 450N ?
 
  • #22
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  • #23
arkofnoah said:
okay... you need to find the perpendicular component of both forces (in green).

http://img843.imageshack.us/img843/5620/screenshot20100806at013.png

and you'd probably need trigo to do that (though I suspect there's some form of symmetry here. hmmm).

Gosh... I appreciate the opportunity u guys had given me to be involved in the thinking procedures of solving this question but i have not learned trigo in my elementary mathematics ... is there an alternative ?
 
Last edited by a moderator:
  • #24
trigo is the surefire and easiest way to do this. i think you might somewhat bypass using trigo by falling back on geometry (from the look of that diagram) but it's probably very clumsy and non-intuitive, if it's possible at all.
 
  • #25
arkofnoah said:
trigo is the surefire and easiest way to do this. i think you might somewhat bypass using trigo by falling back on geometry (from the look of that diagram) but it's probably very clumsy and non-intuitive, if it's possible at all.

Can i take a look at the solution / steps involved??
 
  • #26
Last edited by a moderator:
  • #27
arkofnoah said:
something like this using similar triangles:

http://img689.imageshack.us/img689/4745/screenshot20100806at025.png

just a rough sketch i don't know if i got the alternate angles right so do check it again. you should pick up trigo asap because the above method is rather crude.

So, based on this diagram, what am i supposed to find ?

Can u show me the trigo way of doing it ?
 
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  • #28
based on similar triangles (you learned this before right), find W' (the green one). Assuming that the net moment is zero when F is just enough to balance W, W' = F'. From F' get F using similar triangles again.
 
  • #29
arkofnoah said:
based on similar triangles (you learned this before right), find W' (the green one). Assuming that the net moment is zero when F is just enough to balance W, W' = F'. From F' get F using similar triangles again.

What about the trigo way ??
 
  • #30
OH! actually i just realized we don't need to do all that.

F x 0.3 = 1500 x 0.4

That's it (the sides of the triangle I've drawn is found using pythagoras' theorem). Sorry for complicating the matter :-p
 

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