A see-saw made from 4 meter board sits on a pivot at its center. A child...

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Physics

A see-saw made from 4 meter board sits on a pivot at its center.A child weighing 280 N sits 1.75 m away from the center. A secondchild who weighs 330 N sits on the other side such that the boardremains horizontal and steady. (a) What is the torque about thepivot point due to the weight of the first child? (b) how far fromthe center does the second child sit? (c) What is the force(magnitude and direction) that the pivot applies to the board? Nowyou come and push on the end of the seesaw nearest the second childwith a force of 110 N. You donâ€™t push straight up, but at an angleof 25Â° from the vertical â€“ tilted toward the center of the seesaw.The first child moves to maintain equilibrium. (d) Based on yourintuition, which way should the child move? Towards or away fromthe center? (e) Calculate how far she is now from the center? (f)What is the force now (magnitude and direction) that the pivotapplies to the board?
Please help me with solving (e) and (f) with the equations sothat I can learn how to do it. Thanks

Answer & Explanation Solved by verified expert

4.3 Ratings (851 Votes)

a Torque about the center for child 1 is given as assumingthat the first boy is sitting on right hand sideHere r is the distance from the center and F is the force actingon itThis is in clockwise directionb See Answer

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