Walking The Plank

Sam is standing at one end of a 10 m board on an ice floor. The board’s mass is 30 kg and Sam weighs 50 kg. The ice is assumed to be frictionless, and there is no slipping between Sam and the Plank.

What distance, with respect to the ice on the ground, does Sam travel if he walks to the other end of the board?

None?

The board would move.
It would look like he is walking in place...no? 🤔

@very-musty said
None?

The board would move.
It would look like he is walking in place...no? 🤔

You have the correct idea, but you are missing something.

Possible Clarification: The mass of the board is uniformly distributed over its length.

Still Open

the gravity centre of person + plank does't move. at start that would be 15/13 m from the person. the erson moves t the other side of the lank bringing him 15/13 m the other side of the gravity centre. In tola he traveled 2 x15/13 m or 2.31 m

or something of that kind

@mephisto2 said
the gravity centre of person + plank does't move. at start that would be 15/13 m from the person. the erson moves t the other side of the lank bringing him 15/13 m the other side of the gravity centre. In tola he traveled 2 x15/13 m or 2.31 m

or something of that kind

Close. Check your CM calculation.

It doesn't seem like there is going to be any follow up on this.

As has been noted the center of mass of the system ( man + plank ) remains fixed in position. This is because there are no net external forces on the system in the horizontal direction

We note that initially the center of mass is 1.875 [m] from the leading edge of the board.

     CM = ( 50 [kg] ∙ 0 [m] + 30 [kg] ∙ 5[m] ) / ( 50 [kg]+ 30 [kg] ) = 1.875 [m]

And that is where it stays.

In the final position the man at the other end of the board ) you can calculate the CM and find that it lies behind the man 1.875 [m]

So the man walks to the CM 1.875 [m] and from the CM to the end of the board and additional 1.875 [m], traveling 3.75 [m] relative to the ground.

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The long , formal way:


You can formulate this as a relative motion problem using Newtons 2nd Law.

Let:

𝐹 = Force man applies to the plank
𝑀 = mass of the man
𝑚 = mass of the plank
𝕧 = velocity of the man relative to the ground
𝑣 = velocity of man relative to the plank
𝑝 = velocity of the plank relative to the ground
𝕩 = the position of the man relative to the ground
𝑥 = the position of the man relative to the plank

Looking at free body diagrams of both the man and the plank we can derive the equations of motion:

     𝐹 = 𝑀 ∙ 𝑑𝕧/𝑑t Eq1

     -𝐹 = 𝑚 ∙ 𝑑𝑝/𝑑t Eq2

     𝕧 = 𝑣 + 𝑝 Eq3

By equating Eq1 and Eq2

     𝑀 ∙ 𝑑𝕧/𝑑t + 𝑚 ∙ 𝑑𝑝/𝑑t = 0

Now substitute for 𝑝 in Eq3

     𝑀 ∙ 𝑑/𝑑t( 𝕧 ) + 𝑚 ∙ 𝑑/𝑑t( 𝕧 - 𝑣 ) = 0

This rearranges to:

     𝑑/𝑑t( 𝕧 ) = 𝑚 /( 𝑀 + 𝑚 ) ∙ 𝑑/𝑑t( 𝑣 )

Now eliminate time through the use of the chain rule:

     𝑑𝕧/𝑑t = 𝑑𝕧/𝑑𝑣 ∙ 𝑑𝑣/𝑑t

     𝑑𝕧/𝑑𝑣 = 𝑚 /( 𝑀 + 𝑚 )

Rearrange and bring everything inside the differential we see that the following relationship is implied:

      𝕧 = 𝑚 /( 𝑀 + 𝑚 ) ∙ 𝑣

now begin the transition to position

     𝕧 = 𝑑𝕩/dt

     𝑣 = 𝑑𝑥/dt

Finally we can apply the chain rule once more ( in the same manner as above ) to eliminate time:

     𝑑𝕩 = 𝑚 /( 𝑀 + 𝑚 ) ∙ 𝑑𝑥

Now lets integrate both sides LHS goes from 𝕩 to 0 , and RHS goes from 𝑥 to 0 :

     ∫ 𝑑𝕩 = 𝑚 /( 𝑀 + 𝑚 ) ∙ ∫ 𝑑𝑥

Finally we arrive at:

     𝕩 = 𝑚 /( 𝑀 + 𝑚 ) ∙ 𝑥

Plugging in the relative distance of 10 [m] for 𝑥 ( the man has walked the full length of the plank ) , and all the masses

     𝕩 = 30[kg] /( 50[kg] + 30[kg] ) ∙ 10[m] = 3.75 [m]