Calculating Distance with a Balanced Pivot

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To calculate the distance d for the balanced objects on the pivot, it's essential to consider both translational and rotational equilibrium. The net force (Fnet) should equal zero, indicating that the normal force at the pivot balances the gravitational forces of the masses. Additionally, the torques created by each mass around the pivot must also be equal for rotational equilibrium. Understanding torque involves calculating the weight of each mass multiplied by the perpendicular distance from the center of gravity to the pivot point. A clear sketch is recommended to visualize and determine the necessary distances for the calculations.
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Homework Statement



The two objects in the figure below are balanced on the pivot, with m = 1.8 kg. What is the distance d?

http://i241.photobucket.com/albums/ff4/alg5045/p13-27alt.gif

Homework Equations





The Attempt at a Solution


I haven't attempted a solution because I'm not sure how to set this problem up. I know that a Fnet must be found, and I know that the two masses would be in the equation and I know that the pivot would exert a normal force. I know part of the equation would be Fnet = N(pivot) - (m1 + m2)g, but where do the distances come into play?
 
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aligass2004 said:

Homework Statement



The two objects in the figure below are balanced on the pivot, with m = 1.8 kg. What is the distance d?

http://i241.photobucket.com/albums/ff4/alg5045/p13-27alt.gif

Homework Equations





The Attempt at a Solution


I haven't attempted a solution because I'm not sure how to set this problem up. I know that a Fnet must be found, and I know that the two masses would be in the equation and I know that the pivot would exert a normal force. I know part of the equation would be Fnet = N(pivot) - (m1 + m2)g, but where do the distances come into play?
Your approach will solve for the normal force at the pivot by setting Fnet=0 because the system is in translational equilibrium. But since the loads are balanced, you must also look at the rotational equilibrium about the pivot, wher the sum of the torques must be zero. Are you familiar with torque calculations for moments of the objects about the pivot point?
 
No I'm not.
 
Torque of a mass about a point is just its weight times the perpendicular distance of its cg to the point. The torque of each mass must then balance. You'll have to draw a good sketch to get the proper distances involved.
 
I have no idea how to do that.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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