Calculating Mass in a Canoe Swap: A Physics Problem

[jalapenojam]

Can anyone explain to me how to do the following problem?
Ricardo, mass 75 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 20 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moved 50 cm relative to a submerged log during the exchange and calculates Carmelita's mass, which she has not told him. What is it?

Thanks!

 

[jalapenojam]

i tried using the center of mass for a system of particles formula (m1x1 + m2x2 + ...)/(m1 + m2 + ...) using ricardo's seat as the starting point, but that gave me 3 kg as carmelita's mass, which I'm pretty positive is wrong.

 

[physicsfox]

This question is about the position of the center of mass between the man, the woman and the canoe. Pick an arbitrary x position to measure from, and then determine the center of mass. When they switch positions, the center of mass remains at the same location. The forces that create the movement of the two passengers are considered internal forces and thus do not accelerate (or change the position in this case) of the center of mass.

 

[physicsfox]

It may be easier not to use a spot on the boat as your reference point (since the boat moves). Pick an arbitrary Xo, and then measure everything from that location, both before and after movement.

 

[nevetsnosaj]

i did m1x1 - m3x3 = m2x2 and i got 71.67 as the mass of Carmelita. Is that the right answer?

m1 = Ricardo
m2 = Canoe
m3 = Carmelita

 

[suspenc3]

nevetsnosaj..Can you explain a little more what you did?

 

[nevetsnosaj]

i tried using the center of mass for a system of particles formula (m1x1 + m2x2 + ...)/(m1 + m2 + ...) using ricardo's seat as the starting point, but that gave me 3 kg as carmelita's mass, which I'm pretty positive is wrong.

i'm using his formula and just plug the number in. I think when Ricardo moving, he made the canoe moved for a certain distance. (the lake and the canoe is frictionless so that the momentum that ricardo give to the canoe makes the canoe move) When carmelita move, she make the canoe move to the other way since carmelita is going in opposite direction of Ricardo.
From that thoughts and his formula above, i did what i did.

 

[suspenc3]

so you did m1(x1) - m3(x3) = m2(x2)?

What did you use for the x position of the canoe..ricardo..and carmelita?

 

[nevetsnosaj]

x1 = ricardo = 3m
x2 = canoe = 50 cm
x3 = carmelita = 3m
that's what i did.

 

[Fermat]

Draw two diagrams, one for before the movement of ricardo and carmelita and another diagram for afterwards, as physicsfox suggested.

On the 1st diagram, mark the positions of the mass of ricardo, carmelita and the canoe. Now take moments to find the position of the COM of the canoe plus two canoeists.

When ricardo and carmelita exchange position, this will alter the position of the COM wrt the canoe, (but not the lake) and you know that that difference in position is 50 cm.
Now draw the 2nd diagram, marking the positions of ricardo, carmelita and the canoe again. Take moments again to find the COM of the canoe plus canoeists.

You should now be able to eliminate unknowns.

 

[nevetsnosaj]

so with that we can use the conservation of momentum and eliminate the Velocity and then find the mass of Carmelita?

 

[Fermat]

There is no momentum involved. There are no velocities involved. Simply a change in the COM of the canoe.
The movements of the canoeists are internal forces/actions and have no (external) effect upon the canoe. This means that there is no net force on the canoe/canoeists and hence no acceleration and hence its COM does not change position (wrt to its external environment). But since the COM changes inside the canoe then the canoe must move about in order to keep its original COM in the same postion (wrt the lake).

By taking moments, for the before and after cases, and using the fact that the change in COM is 50 cm, the moment equations involved will be all you need.

 

[nevetsnosaj]

i got confused with the COM and wrt. Isn't COM is Conservation of Momentum?

 

[Fermat]

Sorry, COM = Centre of Mass
wrt = with respect to

 

[nevetsnosaj]

ok. I get it. So since the COM change, the canoe moves.
I've figured out the distance between each person to the COM after the switching. What am I supposed to do with that number?

 

[Fermat]

Well, if you have a (numerical) value for the position of the COM, then the moment eqn will give you the mass of carmelita.

What value do you have ?

 

[nevetsnosaj]

Ricardo to COM = 175 cm
Carmelita to COM = 125 cm

and what's the formula of the moment eqn ?

 

[Fermat]

Hmm.
Ricardo is heavier, so the COM should be closer to him than carmelita.
Good point: The numerical values are correct

 

[nevetsnosaj]

opps... that's a typo... So i get it right, don't i? I don't know what is the moment eqn . Can you tell me the equation?

 

[Fermat]

The moment eqn is the eqn of moments.

Take moments (moment arm - is that term familiar ?) about any point.
e.g. m1*x1 + m2*x2 + ... = 0

Edit: I think you have already done something like this.

 

[nevetsnosaj]

OK so it's going to be:

m1x1 + m2x2 = 0
(75)(125)+(m2)(-175) =0
175m2=9375
m2=53.57 kg

(x2 is negative because m2 is on the other side of m1 and the COM is the zero point)

I've never done it. I'm learning how to do it. Now I understand about the moments eqn. Thanks for the help.

 

[Fermat]

You haven't included the mass of the canoe.

Remember the COM of the canoe doesn't act through the COM of the canoe plus canoeists.

 

[nevetsnosaj]

so we're supposed to to this?

m1x1+m2x2+m3x3=0
(75)(125)+(20)(-50)+(m3)(-175)=0
175m3=8375
m3 = 47.86 kg

 

[Fermat]

Almost there, but the distance between the mass of the canoe and the COM of the whole lot is only 25 cm.

The COM moves from 25cm on one side, to 25 cm on the other side, giving a total movement of 50 cm.

 

[nevetsnosaj]

thanks... then the m3 should be 50.71 from

(75)(125)+(20)(-25)+(m3)(175)=0

 

[Fermat]

Yay! That's it. We finally got there

 

[nevetsnosaj]

thanks... physics is sure a lot of fun.

 

[suspenc3]

how do you find the COM though..Carmelitas mass is unknown

 

[Fermat]

ricardo and carmelita are 3m apart, and the mass of the canoe acts halfway between them. Since ricardo is heavier, the COM, of the canoe plus canoeists, will be on ricardo's side of the distance between him and carmelita. This COM will be at a distance x from the halfway point.
When ricardo and carmelita exchange places, the COM also exchanges places and is now at a distance x from the halfway point, again on ricardo's side.
This means that the COM has moved a distance of 2x.
The question tells you that this distance, 2x, is 50 cm. Ergo the COM is 25 cm from the halfway point.

 

[suspenc3]

Ok...I think That your answer is wrong though.

Heres my thinking...

BEFORE MOVEMENT: $$COMx=m1(-3)+m2(3)+mc(0)$$

AFTER MOVEMENT: $$COMx=m2(-3.5)+m1(2.5)+mc(-0.4)$$

therefore:

$$-3(m1)+3(m2)=-3.5(m2)-0.5(mc)+2.6(m1)$$

$$m2=65.4Kg$$

*im pretty sure that the centre of mass of ricardo+carmelita+canoe does not change when they switch places

 

[Fermat]

...
*im pretty sure that the centre of mass of ricardo+carmelita+canoe does not change when they switch places

The "group" centre of mass does change.
Have you got a ruler handy. Balance it on your finger with two unequal weights (an eraser and a pencil ??) equidistant from the centre. Your finger will be at the COM of the ruler plus two weights. Now switch the positions of those weights. To keep it balanced, your finger will have to move position, so the COM of the "group" has changed.
The total movement of your finger will be twice the distance of it from the mid-point of the ruler.

Edit: could you say where you measured the COM from in each case ?

 

[suspenc3]

I just found this...What do make of it?

http://www.physics.ucc.ie/py1052_ps6.pdf

Q6

 

[suspenc3]

now that I think of it...Their answer makes no sence.

Why would The COM be in the center when the two masses are unequal?

 

[Fermat]

It's a question on conservation of momentum (and energy change). The OP question isn't.

Was there some bit in particular you meant

 

[Fermat]

Ah, you meant Q5.