Plank and torque, problem with finding center mass

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SUMMARY

The discussion focuses on calculating the tipping point of a uniform plank, measuring 5.0 m in length and weighing 212 N, supported at two points with 1.41 m overhanging. The correct torque balance equation is established as 212 N * 1.09 m = 433 N * distance, where the right support serves as the fulcrum. The confusion arises from miscalculating the center of mass, which is located at 2.5 m from either end of the plank, leading to the correct distance of 1.09 m from the right support for torque calculations.

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Homework Statement



A uniform plank of length 5.0 m and weight 212 N rests horizontally on two supports, with d = 1.41 m of the plank hanging over the right support. To what distance, x, can a person who weighs 433 N walk on the overhanging part of the plank before it just begins to tip?

Homework Equations



Tnet = 0
T = r * F

The Attempt at a Solution


I already found the answer, but there's something that bugs me.

Torque of plank = Torque of person
212N * 1.09m = 433N * distance

The right support is the fulcrum point.

What I don't understand is why the plank's weight acts on is 1.09m. I know the idea to get the midpoint from the two supports then subtract by the plank that is hanging over the right support, 5.00m / 2 - 1.41m = 1.09m, but I don't understand why this is the case.

This is what I initially did, 5.00m + 1.41m = 6.41m, then I divided this value by 2 to get 3.205m, and this is the center point of the plank from either end point. This point will be 1.795m from the right support, and I used this to calculate the torque of the plank. Could someone enlighten me why this was wrong? And why the correct way is .. correct?
 
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The beam is 5.0 m in overall length. The distance between supports is 3.59 m. You have misinterpreted the beam as being 6.41 m long and the distance between supports as being 5 m. This is not what is given.
 
ah thank you, I need to read it more clearly.
 
It is a good idea to draw a diagram showing all the forces acting.
The plank is uniform so the weight of 212N is acting at the centre of the plank (2.5m from each end). This should help to see the distancesa from forces and pivots.
 
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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