What forces are exerted on a pivot and its support in a see-saw lever?

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The discussion centers on the forces exerted on a pivot and its support in a see-saw lever scenario. The yellow ball exerts a 1N force and the green ball exerts a 2N force, both positioned 1m from the pivot. The blue bar, which maintains the lever's horizontal position, exerts a force on the red support, while the total force acting down on the pivot is 3N. The upward force from the pivot balances the weights of the yellow and green balls along with the lever itself.

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yhawke
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Hi there,
I have a two part question, just for my understanding not homework. Say you had a see-saw type lever like in my attachment. The yellow ball represents a 1N force and the green 2N and they are both 1m from the pivot. The blue bar is a beam attached hanging down directly off the lever so that it keeps the lever perfectly horizontal. My questions are:

1. What force does the blue bar exert on the red support? (My guess is 1N to the left.)
I think I'm confusing myself because it's directly under the pivot and therefore I'm thinking it has no torque, but surely it is exerting a force?

2. What force is exerted down on the pivot? (My guess is 3N straight down.)
But my thinking is maybe the red support is taking some of the load off the pivot so maybe less than 3N?

Any help is appreciated, I'm an absolute novice trying to teach myself with an old textbook.
 

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got 1) you should know how to play with torque :)

its obvious that the blue and red bar will meet at a point (ad not a surface) ... calculate the distance of this point from the pivot ... the force due to this contact provides restoring torque such that it balances the torques due to the 2 balls.

2) help coming up

EDIT:
2)
for this just use simple Newton's laws ...
upward force due to pivot balances the plank and the yellow and green ball's weights.
 

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