See-saw, force exerted up by pivot?

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Homework Help Overview

The discussion revolves around a physics problem involving a see-saw, where a child exerts force on one end while the other end is secured by a rope. The problem requires determining the total positive force exerted upwards by the pivot, considering the weights involved and the angles of the see-saw.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the calculation of forces and torques acting on the see-saw, questioning the assumptions about the positions of the see-saw and the child. There is discussion about whether the see-saw is in equilibrium and how to account for the forces exerted by the rope and the child.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning and calculations. Some guidance has been offered regarding the calculation of forces, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working under the constraints of the problem's setup, including the see-saw's angle, the weights involved, and the equilibrium condition. There is an emphasis on understanding the relationship between torque and force in this context.

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see-saw, force exerted up by pivot?

A child weighing 36 N climbs onto the right end of a see-saw, little knowing that it is tired down with a rope at its left end. The see-saw is 2.8 m long, weighs 11 N, and is tilted at an angle of 25 degrees from the horizontal. The center of the mass of the see-saw is half way along its length and lies right above the pivot.

There are 5 questions total and I've sucesfully answered 4 of them so far yet this one is odd:

Q: What is the total positive force exerted upwards by the pivot?

I first tried simply adding the two weights together (47 N) but this was wrong; then I added the two forces regarding the torque plus the total weight of objects on the see saw (45.6779 *2 + 36 + 11), but again that was wrong. How do I put these together? thanx.
 
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Is the right end of the seesaw touching the ground? If it was I would be inclined to say that only the weight of the seesaw is on the pivot.
 
lizzyb said:
I first tried simply adding the two weights together (47 N) but this was wrong; then I added the two forces regarding the torque plus the total weight of objects on the see saw (45.6779 *2 + 36 + 11), but again that was wrong. How do I put these together? thanx.

Please explain how you obtained 45.6779 *2. What you should try to calculate is the force that the rope exerts downwards on the see-saw. Can you take it from here?
 
thanks for the responses! the right end (with the child on it) is way up in the air and the rope is holding the left end on the ground.

I came up with 45.6779 * 2 because that is the torque the child exerts on the right hand side and the torque the rope applies (they're the same since they're in equilibrium).
 
oh I see! ok i'll try it.
 
great! thanks! I got it: (36 + 11 + 36 [the rope's force {not torque}])
 
45.6779 * 2. I don't think this is the actual torque, you must have got some calculations mixed up.

You're welcome. Good job in figuring it out.
 

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