Rotanional equilibrium problem..

[411309]

---

Members don't see this ad.

[FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif]A "see-saw" apparatus was set up such that the fulcrum is 3/4 of the length of the uniform bar (x) from one end. A person of mass 2 m sits at the end of the side closer to the fulcrum while another person of mass m sits on the other side of the fulcrum. How far away from the end must the lighter person sit in order for the system to be in equilibrium assuming that the mass of the bar is 1/2 m?
...
[FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif].A. 1/4x
B. 1/2x
C. 3/8x
D.5/16x

The explanation for this was like a page long. I understand how to do these type of problems but this one just kinda confused the crap out of me. Can someone show me how they figured out this problem? Drawing it just confused me even more Lol

.

 

[milski]

There's more than one way to do it. Let's say that the length of the bar is L and we'll consider rotation around the fulcrum (you could pick any point though - the torque around any of them should be zero).

Torque of the heavy person is 2m * L/4.
Torque of the bar itself is 1/2m * -L/4 (the CM of the bar is on the other side of the fulcrum).
Torque of the light person is m * -Lx.

2m * L/4 - 1/2m * L/4 - m * Lx=0
1/2-1/8-x=0
x=1/2-1/8=3/8

or the other person is 3/8L away from the fulcrum.

 

[cloak25]

There's more than one way to do it. Let's say that the length of the bar is L and we'll consider rotation around the fulcrum (you could pick any point though - the torque around any of them should be zero).

Torque of the heavy person is 2m * L/4.
Torque of the bar itself is 1/2m * -L/4 (the CM of the bar is on the other side of the fulcrum).
Torque of the light person is m * Lx.

2m * L/4 - 1/2m * L/4 - m * Lx=0
1/2-1/8-x=0
x=1/2-1/8=3/8

or the other person is 3/8L away from the fulcrum.

Yup. I got the same 🙂.

 

[SaintJude]

How did you know "Torque of the bar itself is 1/2m * -L/4" ?

Edit: P.s.: I now understand why med schools accepts non-traditional students. You guys are on a whole different level...

 

[milski]

How did you know "Torque of the bar itself is 1/2m * -L/4" ?

Edit: P.s.: I now understand why med schools accepts non-traditional students. You guys are on a whole different level...

Torque is arm * force. The force is 1/2mg, because the mass of the bar is 1/2m. The arm is the distance between center of mass (middle of the bar) and the point of rotation (the fulcrum, in this case), which makes it L/4. The sign is negative because it's opposite to the torque of the heavy guy who is on the other side of the fulcrum.

PS: Big correction - all terms that have mass have to be mg instead of m. That does not change the result since m cancels from all of them anyway but it's still incorrect to omit it in the first place.