# Turning effect of forces

1. Aug 5, 2010

### tgpnlyt7095

1. The problem statement, all variables and given/known data

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

2. Relevant equations

n/a

3. The attempt at a solution

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

I think that theres a smaller force, right??

2. Aug 5, 2010

### Staff: Mentor

What's the height of the force F? Where's your pivot?

3. Aug 5, 2010

### tgpnlyt7095

the height is 0.2 + 0.5 = 0.7m.

The pivot , i suppose is the edge of the step ??

4. Aug 5, 2010

### arkofnoah

Last edited by a moderator: May 4, 2017
5. Aug 5, 2010

### tgpnlyt7095

thanks :D

Last edited by a moderator: May 4, 2017
6. Aug 5, 2010

### Staff: Mentor

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?

7. Aug 5, 2010

### tgpnlyt7095

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

8. Aug 5, 2010

### Staff: Mentor

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?

9. Aug 5, 2010

### tgpnlyt7095

It should be, F x D = 1500N x 0.7m = 1050Nm

10. Aug 5, 2010

### Staff: Mentor

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. Aug 5, 2010

### tgpnlyt7095

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. Aug 5, 2010

### arkofnoah

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. Aug 5, 2010

### tgpnlyt7095

But i thought its a circle ??

14. Aug 5, 2010

### tgpnlyt7095

I think i got it.

F x D , = 1500N x ( 0.5m - 0.2m )
= 450Nm

15. Aug 5, 2010

### Staff: Mentor

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?

16. Aug 5, 2010

### tgpnlyt7095

0.5m - 0.2m = 0.3m

17. Aug 5, 2010

### arkofnoah

actually the distance of Barrel from the centre to the edge of the step is the radius of the circle :tongue:

18. Aug 5, 2010

### Staff: Mentor

LOL... yeah, I should have said "What is the horizontal distance from barrel center to the edge of the step".

19. Aug 5, 2010

### tgpnlyt7095

Sorry .. ;(

20. Aug 5, 2010

### Staff: Mentor

What's the height of F? What's the height of the step? The vertical distance between them is D.