Amount of work in displacing a brick

[rudransh verma]

Homework Statement

A 2kg brick of dimension 5m*2.5m*1.5m is lying on the largest base. It is now made to stand with length vertical, then amount of work done is (g=10)

Relevant Equations

W=F.d

W=-mgd=2*10*1.5=30J. I don’t think this is right.

 

[Orodruin]

What is your reasoning and why do you think it is wrong?

 

[haruspex]

Did you draw the two diagrams?

 

[vcsharp2003]

Homework Statement:: A 2kg brick of dimension 5m*2.5m*1.5m is lying on the largest base. It is now made to stand with length vertical, then amount of work done is (g=10)
Relevant Equations:: W=F.d

W=-mgd=2*10*1.5=30J. I don’t think this is right.

Wouldn't length be 5m? In your calculation you have taken length to be 1.5m. My understanding is that the brick is initially sitting on ground with the face 5m X 2.5m in contact with the ground.

 

[Delta2]

Wouldn't length be 5m

Nope according to my opinion. The brick is not a single point it has dimensions. We 've got to consider the COM of the brick (for which we need to know its mass distribution except the dimensions and shape, but we can assume that its mass is homogenously distributed). We 've got to take the difference in the height of the COM of the brick in the two positions to correctly compute the work.

 

[vcsharp2003]

Nope according to my opinion. The brick is not a single point it has dimensions. We 've got to consider the COM of the brick (for which we need to know its mass distribution except the dimensions and shape, but we can assume that its mass is homogenously distributed). We 've got to take the difference in the height of the COM of the brick in the two positions to correctly compute the work.

What I meant was that the brick is moved/rotated so it's 5 m side is standing vertical in its final position since question states that "It is now made to stand with length vertical, ".

 

[Delta2]

What I meant was that the brick is moved/rotated so it's 5 m side is standing vertical in its final position since question states that "It is now made to stand with length vertical, ".

Ah ok, I thought you meant that we should take the $d$ in @rudransh verma formula to be 5 instead of 1.5.

 

[rudransh verma]

Nope according to my opinion. The brick is not a single point it has dimensions. We 've got to consider the COM of the brick (for which we need to know its mass distribution except the dimensions and shape, but we can assume that its mass is homogenously distributed). We 've got to take the difference in the height of the COM of the brick in the two positions to correctly compute the work.

Frankly I don’t know how to do rotation problems since rotation chapter has yet to come in future. That’s a question from work, energy and circular motion. Can we solve it without COM or anything like that?

 

[Delta2]

Frankly I don’t know how to do rotation problems since rotation chapter has yet to come in future. That’s a question from work, energy and circular motion. Can we solve it without COM or anything like that?

The easy way is to do it with the COM, and we bypass all the details about the rotation and translation of the body since the work done by the weight (which we consider to be concentrated at the COM of the body), which is a conservative force depends only on the initial and final position and not in the inbetween path that it takes. So we find the initial height of the COM, the final height of the COM, we take the difference and plug it in that formula and we are finished.

The hard way is to take into account how exactly the body travels form the initial position to the final position, which depends on how exactly it is rotated and translated, and then compute a hard line integral. Why your mind goes to the hard way, this is an introductory problem, think there should be an easy way :D.

 

[Frabjous]

The first thing you need to do is calculate the potential energy due to gravity of the brick. You do not need to use “center of mass” to do this. Show us how you do this. Do not just give us an answer.

 

[Delta2]

The first thing you need to do is calculate the potential energy due to gravity of the brick. You do not need to use “center of mass” to do this. Show us how you do this. Do not just give us an answer.

Sorry I don't think one can calculate the potential energy of a rigid body in a gravitational field without knowing the position of the COM (at least not in an easy way) but I am all ears.

 

[Orodruin]

Ah ok, I thought you meant that we should take the $d$ in @rudransh verma formula to be 5 instead of 1.5.

No. And may I remind everyone of the homework helper rules not to solve the problems before the OP has.

 

[rudransh verma]

Why your mind goes to the hard way, this is an introductory problem, think there should be an easy way :D.

You got me. This is the problem I have been struggling for years or maybe it’s how my mind works. Sad

The first thing you need to do is calculate the potential energy due to gravity of the brick.

As far as I know brick is always on the ground. So it’s U=0.

 

[Frabjous]

Sorry I don't think one can calculate the potential energy of a rigid body in a gravitational field without knowing the position of the COM (at least not in an easy way) but I am all ears.

It’s a brick lying flat on one of it’s surfaces for both configurations in the problem. It is a simple calculation.

 

[Frabjous]

As far as I know brick is always on the ground. So it’s U=0.

Do you believe that both positions of the brick described in the problem have the same potential energy?

 

[BvU]

You got me. This is the problem I have been struggling for years or maybe it’s how my mind works. Sad

As far as I know brick is always on the ground. So it’s U=0.

Make two drawings

 

[rudransh verma]

Make two drawings

 

[Steve4Physics]

This (surprisingly large) brick measures 5m x 2.5m x 1.5m. And its mass is only 2kg.

That means its density is far less than the density of air. So the brick will just float away. They don’t make bricks like they used to.

 

[BvU]

Good. Now: how far do you have to lift the left one, in order to get it on the same average height as the right one?

 

[Delta2]

It’s a brick lying flat on one of it’s surfaces for both configurations in the problem. It is a simple calculation.

I don't think it's simple if you don't use the center of mass (and assume that the brick mass density is homogeneously distributed). PM please how you do this simple calculation without the notion of COM.

 

[rudransh verma]

Good. Now: how far do you have to lift the left one, in order to get it on the same average height as the right one?

I guess $\sqrt{5^2+5^2}=5\sqrt2$

 

[BvU]

This is PF, not a guessing game. How did you come to this strange result?

 

[BvU]

Draw the 'lifted' case according to your result...

 

[rudransh verma]

How did you come to this strange result?

Like this. I took a corner and calculated how far that corner travels but for different corners its different.

 

[BvU]

But if you lift the whole block by that much, it is well above the other!

 

[Delta2]

That's why I said you got to take the position of the COM , not that of a corner, and find not exactly how much the COM travels but the difference in height.

 

[Frabjous]

This approach only works using the idea of center of mass, which is why I suggested calculating the potential energy directly.

 

[BvU]

Like this. I took a corner and calculated how far that corner travels but for different corners its different.

Lifting means moving straight upwards...

 

[rudransh verma]

the COM travels but the difference in height.

W=-mg(2.5-0.75)=-2*10*1.75=-35 J

This approach only works using the idea of center of mass, which is why I suggested calculating the potential energy directly.

The brick is always on the ground. So its U=0J

 

[Delta2]

W=-mg(2.5-0.75)=-2*10*1.75=-35 J

Now you got it right I believe but do you understand why we take the position of COM and why only the difference in height of the COM is what matters? (and not for example how much it travels in straight line or in a curved trajectory)

The brick is always on the ground. So its U=0J

This argument is too much casual logic and not scientific. One surface of the brick is on the ground and not the whole brick (in any of the two positions).