Suppose in the previous problem m3 balances the see-saw by pulling now

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Homework Statement



Suppose in the previous problem m3
balances the see-saw by pulling now on the end 1.5 m from the ful-crum at an angle of 30° from the horizontal as shown.
What force is necessary for balance?

Homework Equations



See picture for both questions
The first question i got 31.4 kg

second question it added the angle and i don't know what to do with that.

Also I don't understand why does the mass of the seesaw doesn't matter?

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oneshot said:

Homework Statement



Suppose in the previous problem m3
balances the see-saw by pulling now on the end 1.5 m from the ful-crum at an angle of 30° from the horizontal as shown.
What force is necessary for balance?

Homework Equations



See picture for both questions
The first question i got 31.4 kg
Yes, looks good
second question it added the angle and i don't know what to do with that.
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.
Also I don't understand why does the mass of the seesaw doesn't matter?
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?
 
I actually don't understand how does the breaking component work...
Fx cos 30 + Fy sin 30?
 
oneshot said:
I actually don't understand how does the breaking component work...
Fx cos 30 + Fy sin 30?
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.
 
PhanthomJay said:
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.


oh make sense. but u said ignore m3? why? torque = rF , so the force does not include m3g?
 
oneshot said:
oh make sense. but u said ignore m3? why? torque = rF , so the force does not include m3g?
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.
 
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.


oh ok thanks!