How Far Can the Painter Walk Without Tipping the Plank?

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The discussion focuses on a physics problem involving a painter walking on a plank balanced on sawhorses, with specific weights and dimensions provided. Participants express confusion about how to calculate the tipping point of the plank, particularly regarding the forces acting on it and the placement of the paint bucket. Key points include the need to balance forces and torques to determine how far the painter can walk without tipping the plank. There is a suggestion to consider the center of mass for the paint bucket and to choose an appropriate rotational axis for torque calculations. The conversation emphasizes the importance of understanding the mechanics involved to arrive at the correct solution.
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Homework Statement



A painter (mass 68 kg) is walking along a trestle, consisting of a uniform plank (mass 21.0 kg, length 6.00 m) balanced on two sawhorses. Each sawhorse is placed 1.40 m from an end of the plank. A paint bucket (mass 7.0 kg, diameter 28 cm) is placed as close as possible to the right-hand edge of the plank while still having the whole bucket in contact with the plank.

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(a) How close to the right-hand edge of the plank can the painter walk before tipping the plank and spilling the paint?
______ m

(b) How close to the left-hand edge can the same painter walk before causing the plank to tip? (Hint: As the painter walks toward the right-hand edge of the plank and the plank starts to tip clockwise, consider the force acting upward on the plank from the left-hand sawhorse support.)
______ m



Homework Equations





The Attempt at a Solution



I am just completely confused on this problem. I don't even know where to start.
 
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You shouldn't have any problem finding the mg forces downward of the painter, bucket and board and their positions. You have two unknown upward forces exerted by the sawhorses. But these two unknown forces can be computed by balancing the sum of the forces and sum of the torques. When one of the upward forces vanishes, you are on the edge of tipping, because the sawhorses can't pull down.
 
what do i do about the can at the end when it says its diameter is .28 meters? where do i take the R from?
 
Put the force at the center of mass. .14m from the end of the board. Ditto for all others.
 
ok I did that and it still doesn't give me the correct answer.

for the balancing of torques, do you imput the R as the distance from the rotational axis?
 
To know why the answer isn't correct we'd have to know a lot more about what you did. You can put the presumed rotational axis anywhere you want. If it doesn't rotate around one axis, then it doesn't rotate around any. It's usually convenient to put it at the location of one of the forces. Then you don't have to put that force into the torque equation.
 
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