See-saw, force exerted up by pivot?

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The discussion revolves around calculating the total positive force exerted upwards by the pivot of a see-saw with a child on one end and a rope securing the other. A child weighing 36 N is on the right end, while the see-saw itself weighs 11 N and is tilted at 25 degrees. Initial attempts to find the force involved adding weights incorrectly, leading to confusion about torque and equilibrium. The correct approach involves considering the downward force exerted by the rope and the weights on the see-saw. Ultimately, the participants clarify the calculations needed to determine the upward force at the pivot.
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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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