Shift in position of plank when a man moves on it (Centre of mass)

[ShaunPereira]

Homework Statement

A uniform wooden plank of mass 150 kg and length 8 m is floating on still water with a man of 50 kg at one end of it . The man walks to the other end of the plank and stops. Than the distance covered by the plank is

Relevant Equations

X(com)=(m1x1 + m2x2)/m1+m2

Lets take the original position of the man to be our origin
The plank is uniform so we can assume its mass to be concentrated at its center i.e. 4m from the origin

X_com= m₁x₁+m₂x₂/m₁+m₂
=50(0) +150(4) /50+150
=3m

There is no external force on the system so the centre of mass does not move
The man moves to the other side of the plank so the distance form origin becomes 8m
To compensate for this the plank moves a distance d. The original position of the centre of plank was 4m from the origin so its new distance from the origin is (4-d)

X_com= 3 = 50(8)+ 150(4-d) / 150+50
this gives us d=8/3 m

The answer is incorrect according to the solution provided to me
It should be 2m according to it

 

[Orodruin]

The man moves to the other side of the plank so the distance form origin becomes 8m

If the man moves 8 m and the plank moves 8/3 m in the opposite direction, the man will be very wet.

 

[ShaunPereira]

If the man moves 8 m and the plank moves 8/3 m in the opposite direction, the man will be very wet.

Indeed

so if the boat moves a distance d the man moves a distance d too since he is on the boat
the equation becomes

3= (4-d)150 + (8-d)50 /150+50
this gives us d=2m

Is this correct?

 

[PeroK]

Homework Statement:: A uniform wooden plank of mass 150 kg and length 8 m is floating on still water with a man of 50 kg at one end of it . The man walks to the other end of the plank and stops. Than the distance covered by the plank is
Relevant Equations:: X(com)=(m1x1 + m2x2)/m1+m2

Lets take the original position of the man to be our origin
The plank is uniform so we can assume its mass to be concentrated at its center i.e. 4m from the origin

X_com= m₁x₁+m₂x₂/m₁+m₂
=50(0) +150(4) /50+150
=3m

Correct.

There is no external force on the system so the centre of mass does not move

Correct.

The man moves to the other side of the plank so the distance form origin becomes 8m

This is not right, as the plank moves.

To compensate for this the plank moves a distance d. The original position of the centre of plank was 4m from the origin so its new distance from the origin is (4-d)

Okay.

so if the boat moves a distance d the man moves a distance d too since he is on the boat

I think you mean the man moves a distance $8-d$. And plank not boat.

the equation becomes

3= (4-d)150 + (8-d)50 /150+50
this gives us d=2m

Is this correct?

Yes.

Note that as the plank is three times the mass of the man, the man moves three times as far, so $d + 3d = 8m$ and $d = 2m$.

 

[ShaunPereira]

I think you mean the man moves a distance 8−d. And plank not boat.

Yes. I meant to say that the man moves 8 m say left but the boat moves a distance d towards right .The man is on the boat and hence also moves a distance d towards right otherwise he would fall into the water as Orodruin rightly pointed out.

so the net distance that the man has traveled from the origin becomes 8-d

 

[ShaunPereira]

Yes.

Note that as the plank is three times the mass of the man, the man moves three times as far, so d+3d=8m and d=2m.

Ok. Thank you!

 

[PeroK]

Ok. Thank you!

Do you see why?

 

[ShaunPereira]

Do you see why?

Well I thought that was a useful observation. Since the plank is three times heavier than the man, whatever distance the plank displaces, the man being three times lighter would have to cover thrice the distance to balance out the displacement of the plank.

I reckon this analogy could be used even when we change the masses of the man and the plank. Say the plank is 5 times heavier than the man therefore whatever distance the plank is displaced by the man would have to cover 5 times that

We could thus formulate an expression for this
m1Δx1 =m2Δx2

where Δx = net displacement
man moves 8m to the left. Boat moves Δx2 to right
therefore net displacement of man is Δx1= (8-Δx2)

m1(8-Δx2) =m2Δx2
Here m2=5m1

On solving we get
Δx2= 4/3 m This is the net displacement of the boat
Substituting this value in the original formula we made
m1Δx1 = m2Δx2 we get

Δx1= 20/3 m This is the net distance the man is displaced from the origin.

Is this correct?

 

[PeroK]

Yes, precisely. Technically, of course, the displacements are in opposite directions.

 

[ShaunPereira]

the displacements are equal in opposite directions.

Hmmm... I didn't get you there. How can the displacements be equal? They are after all dependent on the masses of the two bodies?

 

[PeroK]

Hmmm... I didn't get you there. How can the displacements be equal? They are after all dependent on the masses of the two bodies?

Just a typo!

 

[ShaunPereira]

Just a typo!

OK. Thanks!