# How Far Can a Painter Walk on a Plank Before It Tips?

• Toranc3
In summary, the painter can walk close to the right-hand edge of the plank before causing it to tip, but not close to the left-hand edge.
Toranc3

## Homework Statement

A painter (mass 61 kg) is walking along a trestle, consisting of a uniform plank (mass 20.0 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 4.0 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.

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

## Homework Equations

Summation of forces

Summation of torques , tao=Fr

## The Attempt at a Solution

Wp=weight of the plank, wg= weight of the guy, wb weight of the bucket, nr=normal of the right hand fulcrom

Wp=196N, Wg=588N, Wb=39.2N

Fy: -Wp-Wg-Wb+Nr
Nr=823.2 Newtons

Torque: N(1.6m)-Wg*x

x=2.24 meters.

I did my torque about the center of the plank. Wp

this is for part a. i am still working on part b and i will post it up in a bit. So basically in the torque you exclude the normal on the left hand fulcrum and the weight of the paint. is this right?

[url=http://www.freeimagehosting.net/pq8zg][PLAIN]http://www.freeimagehosting.net/t/pq8zg.gif[/url][/PLAIN]

Last edited:
drawing is now up.forgot to post it up so sorry.

Why would you exclude the weight of the paint? Why do you exclude the left hand support reaction? And from what point is 'x' being measured?

Take your torques around the point that you expect the plank to pivot. Don't ignore any of the masses on either side of the pivot point; As far as torques are concerned, the weights act through the center of mass for each object (or portion of the object producing a given torque) in question.

PhanthomJay said:
Why would you exclude the weight of the paint? Why do you exclude the left hand support reaction? And from what point is 'x' being measured?

my origin is all the way to the left end of the plank, sorry. I also thought that the sentence meant that when the normal on the left is zero and the paint has fallen off. It is implying only when the normal on the left is zero only though right?

gneill said:
Take your torques around the point that you expect the plank to pivot. Don't ignore any of the masses on either side of the pivot point; As far as torques are concerned, the weights act through the center of mass for each object (or portion of the object producing a given torque) in question.

I redid it and got this:

Wp=weight of the plank, wg= weight of the guy, wb weight of the bucket, nr=normal of the right hand fulcrom

Wp=196N, Wg=588N, Wb=39.2N

Fy: -Wp-Wg-Wb+Nr
Nr=823.2 Newtons

Torque(from Nr): Wp(1.6m)-Wg(x)-Wb(2.86m)

588N(x)=313.6N*m-112.112N*m
x=0.343m from Nr

Toranc3 said:
my origin is all the way to the left end of the plank, sorry. I also thought that the sentence meant that when the normal on the left is zero and the paint has fallen off. It is implying only when the normal on the left is zero only though right?
Yes, correct.

Toranc3 said:
I redid it and got this:

Wp=weight of the plank, wg= weight of the guy, wb weight of the bucket, nr=normal of the right hand fulcrom

Wp=196N, Wg=588N, Wb=39.2N

Fy: -Wp-Wg-Wb+Nr
Nr=823.2 Newtons

Torque(from Nr):0 = Wp(1.6m)-Wg(x)-Wb(2.86m)
The distance from the bucket to the right pivot is not 2.86m.
588N(x)=313.6N*m-112.112N*m
make correction for moment caused by bucket about right pivot
x=0.343m from Nr
when you correct this value for x, it is measured from the right pivot. The problem asks for the distance from the right edge of the plank.

PhanthomJay said:
Yes, correct.

The distance from the bucket to the right pivot is not 2.86m. make correction for moment caused by bucket about right pivotwhen you correct this value for x, it is measured from the right pivot. The problem asks for the distance from the right edge of the plank.

I messed uo again lol. Here is my fix:

Torque: -Wg*x-Wb(1.26m)+Wp(1.6m)

-588N*x-39.2N(1.26m)+196N(1.6m)
x=.449m

So from Nr to him is 0.449m But from the right end to him is 1.4m-0.449m. Would this be correct?

Last edited:
Toranc3 said:
If I understand you correctly. I should make my torque about the right end.

Here is my new work. Forces are all the same as above.

Torque: Wp(3m)-Nr(1.4m)+Wg(x)+Wb(.14m)
after multiplying
588-823.2(1.4)+5.488=-588x

x=.391m0.95 m
You can sum moments about any point you want, and the distance x is the distance from the point about which you sum moments. So your last approach summing momens about the right edge is OK, once you make the correction I noted in red above. But as gneill pointed out, it is better to sum the moments about the right support, since this avoids the need to calculate the value of N_r. If you were to do it this way, you would get a value for x measured from the right support. Then to get the value from the right edge, it would be (1.4 - x).

PhanthomJay said:
You can sum moments about any point you want, and the distance x is the distance from the point about which you sum moments. So your last approach summing momens about the right edge is OK, once you make the correction I noted in red above. But as gneill pointed out, it is better to sum the moments about the right support, since this avoids the need to calculate the value of N_r. If you were to do it this way, you would get a value for x measured from the right support. Then to get the value from the right edge, it would be (1.4 - x).

Hey i just fixed it! I thought you could only do torques about a force. I did not know that you could do it on the right end and that is why I changed it.

PhanthomJay said:
You can sum moments about any point you want, and the distance x is the distance from the point about which you sum moments. So your last approach summing momens about the right edge is OK, once you make the correction I noted in red above. But as gneill pointed out, it is better to sum the moments about the right support, since this avoids the need to calculate the value of N_r. If you were to do it this way, you would get a value for x measured from the right support. Then to get the value from the right edge, it would be (1.4 - x).

So it is .951m then. Alright thanks. Let me post up the next one. :)

Hey sup. Here is part b:

Same y forces.

Torque(about Nl) nl is left fulcrum.

Torque: Wg(x)-Wp(1.6m)-Wb(4.46m)

-Wg*x=-Wp(1.6m)-Wb(4.46m)

-588N*x=-190N(1.6m)-39.2N(4.46m)

x=0.814 meters. from fulcrum

from left end. 1.4m-0.814m=0.586 meters.
Is this correct?

Toranc3 said:
Hey sup. Here is part b:Same y forces.

Torque(about Nl) nl is left fulcrum.

Torque: Wg(x)-Wp(1.6m)-Wb(4.46m)that 4.46 should be 4.26, right?

-Wg*x=-Wp(1.6m)-Wb(4.46m)

-588N*x=-190N196?(1.6m)-39.2N(4.46m)

x=0.814 meters. from fulcrum

from left end. 1.4m-0.814m=0.586 meters.
Is this correct?
More or less, with math/typo corrections, then this should be OK, good!

PhanthomJay said:
More or less, with math/typo corrections, then this should be OK, good!

Is each flucrum equidistant from the center of the plank?. I keep getting 4.46m.

From Left fulcrum to center is 1.6m, then from center to right fulcrum is 1.6m too, then add 1.4m-.14m. That is what I did. Is that wrong?

4.46m looks right to me, but Wp is indeed 196N. Btw, it would have been slightly simpler to work in terms of multiples of g rather than converting to Newtons. All the g's would have cancelled.

haruspex said:
4.46m looks right to me, but Wp is indeed 196N. Btw, it would have been slightly simpler to work in terms of multiples of g rather than converting to Newtons. All the g's would have cancelled.

Alright thanks to both of you. :)

## What is static equilibrium?

Static equilibrium is a state in which an object is at rest and has no net forces acting upon it. This means that all the forces acting on the object are balanced and cancel each other out.

## How is static equilibrium different from dynamic equilibrium?

Static equilibrium refers to a state of rest, while dynamic equilibrium refers to a state of motion where an object is moving at a constant velocity. In dynamic equilibrium, the net force on an object is still equal to zero, but the object is in motion.

## What is the equation for calculating static equilibrium?

The equation for calculating static equilibrium is ΣF = 0, which means the sum of all forces acting on an object must equal zero for it to be in static equilibrium.

## What are some real-life examples of static equilibrium?

Some real-life examples of static equilibrium include a book resting on a table, a person standing on a scale, and a ladder leaning against a wall. In each of these examples, the forces acting on the object are balanced and the object remains at rest.

## How can I determine if an object is in static equilibrium?

An object is in static equilibrium if it is at rest and has no acceleration. To determine if an object is in static equilibrium, you can draw a free body diagram and calculate the net force on the object. If the net force is equal to zero, the object is in static equilibrium.

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