Where Should the Pivot Be Placed to Balance a See-Saw?

[panda02]

Homework Statement

If Sara (100kg) and Jim (200kg) are 1m from the edge of opposite sides of a 13m long, 100kg see-saw, where does the pivot point need to be in order to balance the see-saw?

Relevant Equations

Ts=Tj

Torque_left = 100 kg * 1 m = 100 kg·m
Torque_right = 200 kg * (13 - x) m = 200(13 - x) kg·m
100 kg·m = 200(13 - x) kg·m
100 = 200(13 - x)
x = 12.5 meters

pivot = 12.5 m

 

[Hill]

Homework Statement: If Sara (100kg) and Jim (200kg) are 1m from the edge of opposite sides of a 13m long, 100kg see-saw, where does the pivot point need to be in order to balance the see-saw?
Relevant Equations: Ts=Tj

Torque_left = 100 kg * 1 m

It should be 100 kg times the distance from the pivot point, but 1 m is the distance from the edge.

 

[Hill]

Please make the title of your thread more informative. We already know that it is in the Homework Help forum.

 

[berkeman]

Please make the title of your thread more informative. We already know that it is in the Homework Help forum.

Looks like the OP has fixed it up a little now. Thanks.

 

[kuruman]

Homework Statement: If Sara (100kg) and Jim (200kg) are 1m from the edge of opposite sides of a 13m long, 100kg see-saw, where does the pivot point need to be in order to balance the see-saw?
Relevant Equations: Ts=Tj

Torque_left = 100 kg * 1 m = 100 kg·m

How do you know that Sara is 1 m away from the pivot point?

 

[BvU]

Homework Statement: If Sara (100kg) and Jim (200kg) are 1m from the edge of opposite sides of a 13m long, 100kg see-saw, where does the pivot point need to be in order to balance the see-saw?
Relevant Equations: Ts=Tj

Torque_left = 100 kg * 1 m = 100 kg·m
Torque_right = 200 kg * (13 - x) m = 200(13 - x) kg·m
100 kg·m = 200(13 - x) kg·m
100 = 200(13 - x)
x = 12.5 meters

pivot = 12.5 m

Hello @panda02 ,

$\qquad$ !​

hehe, at least this thread got out of the starting block.

I advise you to stop using torque_left and _right. Just torque and a position for the axis of rotation..

Torque contributions are positive if going counterclockwise, negative if clockwise.

And a picture ! (edit: updated pic)

Force balance: The support has to compensate 400 g
Torque:
If we choose the left end of the board as axis of rotation, we have
from Sara $\ \ -$ 100 g x 1 m
from the board weight $\ \ -$ 100 g x 6.5 m
from Jim $\ \ -$ 200 g x 12 m
and from the supporting pivot point $\ \ +$ 400 g x X m

with X the distance from the left end of the board to the pivot point.

$\$