# Homework Help: Spool problem

1. Nov 29, 2006

### Fusilli_Jerry89

1. The problem statement, all variables and given/known data
A 17N spool has an outside radius of 20 cm and an inside radius of 6.4cm as shown. What horizontal force on the rope(wrapped around the inside radius) will be required to pull the spool over an 11cm edge?
http://img299.imageshack.us/img299/6105/27kl1.png [Broken]

2. Relevant equations
FdL+/-FdL=0
F=force
d=distance to pivot
L=perpendicular

3. The attempt at a solution
0.154F-(17)(0.09)
F=10N
Is this right?

Last edited by a moderator: May 2, 2017
2. Nov 29, 2006

### Fusilli_Jerry89

Can any one help?

3. Nov 29, 2006

### Staff: Mentor

4. Nov 29, 2006

### Staff: Mentor

No. Where did you get the 0.09 m distance? (Remember: You need the perpendicular distance to the pivot point. Perpendicular to what?)

5. Nov 29, 2006

### Fusilli_Jerry89

wait instead of 0.09 should it be 0.20? I'm using the corner as the pivot point.

6. Nov 29, 2006

### Staff: Mentor

Describe to me what distance you are trying to specify: The perpendicular distance from the corner to what? What's the orientation of that distance: horizontal, vertical, at some angle?

7. Nov 29, 2006

### Fusilli_Jerry89

the 0.20 is the horizontal distance from the middle of the spool to the edge. Just the radius.

8. Nov 30, 2006

### Staff: Mentor

Since you are trying to find the torque exerted about the corner edge by the spool's weight, what you need is the perpendicular distance from the corner to the line of the force. Since the weight is vertical, the perpendicular distance will be the horizontal distance between the corner and the center of the spool. That distance is not 0.20 m, although it may look like that since your diagram is not drawn to scale. 0.20 m is the radius of the spool, not the distance to the corner edge.

9. Nov 30, 2006

### Fusilli_Jerry89

ok I see how the distance would get less and less as you go down the circle. But Ihave no idea how to calculate it?

10. Nov 30, 2006

### OlderDan

Draw a radius from the center of the spool vertically downward. Draw another radius to the point of contact with the step. What is the angle between those radii?

11. Nov 30, 2006

### Fusilli_Jerry89

with that triangle i got 58 degrees from the origin and then two 61 degrees. Ur asking for that 58 degrees right?

12. Nov 30, 2006

### Fusilli_Jerry89

k I solved and got approx 0.17 instead of the 0.20 I said earlier: so instead would it be:

0.154F=(17)(0.17)
F=19N

13. Dec 1, 2006

### OlderDan

I got something a bit different for the angle. The center of the circle is 9cm above the contact point. The angle between a horizontal radius and the radius to the contact point is sin^-1(9/20) = 26.7° so the angle to the vertical is 63.3°. You can use either angle to get the distance between the line of gravity and the contact point as 17.9cm. That's not very different from what you got. The rest looks good.