Yes, looks goodoneshot said:
forget about the mass of the person m3 and just consider a pull force as shown. Find its torque about the pivot by breaking it into its components, then proceed as before.
it is assumed that the see saw is uniform and its mass acts at the center (at the pivot), so if you use the pivot as the point about which to sum moments equal 0, does it make a difference?
If you denote the pulling force as F, then the horizontal x component is Fcos30 and the vertical y component is Fsin30. The x component does not produce any torque about the pivot, since there is no perpendicular moment arm. Only the y component does produce torque.oneshot said:
PhanthomJay said:
The problem isn't worded that good, but presumably, the person called 'm3' is standing on the ground as (s)he pulls, so his or her weight does not act on the seesaw (draw a free body diagram of the seesaw), only the pulling force F acts on it.oneshot said:
PhanthomJay said:
m3 balanced the see-saw by pulling on one side with a force that was equal in magnitude and opposite in direction to the combined forces of the other objects on the see-saw.
When m3 pulls on the see-saw, it exerts a force on the see-saw in a specific direction. This force can be measured in units of Newtons (N).
Before m3 pulled, there were forces acting on the see-saw from the other objects present. These forces may have been unbalanced, causing the see-saw to be tilted in one direction.
m3's force on the see-saw causes an equal and opposite reaction force on m3. This reaction force may cause the see-saw to move in the opposite direction, depending on the mass and position of the other objects on the see-saw.
It is possible for m3's force to balance the see-saw on its own, but it depends on the relative masses and positions of the other objects on the see-saw. If the other objects are heavier or positioned closer to the fulcrum, m3 may need to exert a greater force to balance the see-saw.
