Acceleration of center of mass of cart-block system

[esoteric deviance]

http://www.fileden.com/files/2006/11/15/381656/My%20Documents/problem%2013.jpg
The figure above shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block.
The cart has mass m₁ = 0.600 kg, and its center is initially at xy coordinates (-0.500 m, 0 m).
The block has mass m₂ = 0.400 kg, and its center is initially at xy coordinates (0, 0.100 m).
The cart is released from rest, and both cart and block move until the cart hits the pulley.
The friction between the cart and air track and between the pulley and its axle is negligible as is the mass of the cord and pully.

• In unit-vector notation, what is the acceleration of the center of mass of the cart-block system?
• What is the velocity of the com as a function of time t?

so..i know that a_com = F_net/M and that i probably have to find a_com,x and a_com,y, but I'm not sure and if i do need to solve for those, how do i find the F_net?

 

[Dorothy Weglend]

I think you are making it too complicated. The internal forces will cancel, by the 3rd law, right? Fnet would just be the external forces.

Dorothy

 

[esoteric deviance]

Still not sure what to do, sorry :(.

 

[Dorothy Weglend]

The only forces are the tension on the cord, and gravity, as far as I can tell. The tension on the cord is an internal force, which just cancels out by Newton's third law. So Fnet has got to have something to do with gravity, it would seem.

 

[esoteric deviance]

so if the only external force acting on the system is gravity, then
F_net,system = Mg = (m₁ + m₂)g = 9.8?

 

[OlderDan]

so if the only external force acting on the system is gravity, then
F_net,system = Mg = (m₁ + m₂)g = 9.8?

There is another external force besides gravity. You don't need to find it to do the problem, but you could. It is the force of the pulley acting on the string. You can continue to find the motions of the individual masses and then find the motion of the CM from those. But you have too many g in your equation, and something is missing.

 

[esoteric deviance]

Yeah, I guess I'm pretty lost. For some reason, the more I try to think about this problem, the more confused I seem to become :(.
I hate to ask this, but might you be willing to give me a step-by-step explanation of how to solve it?

I don't know if they have any significance (or if they are even correct), but here are some values I've come up with so far:

• x₁ = -0.5, y₁ = 0
• x₂ = 0, y₂ = -0.1
• x_com,sys = (m₁x₁ + m₂x₂)/(M) = -0.3
• y_com,sys = (m₁y₁ + m₂y₂)/(M) = -0.04

and a couple of the equations I've been trying to work with:

• F_net = Ma_com
• Ma_com = m₁a₁ + m₂a₂

 

[OlderDan]

Yeah, I guess I'm pretty lost. For some reason, the more I try to think about this problem, the more confused I seem to become :(.
I hate to ask this, but might you be willing to give me a step-by-step explanation of how to solve it?

I don't know if they have any significance (or if they are even correct), but here are some values I've come up with so far:

• x₁ = -0.5, y₁ = 0
• x₂ = 0, y₂ = -0.1
• x_com,sys = (m₁x₁ + m₂x₂)/(M) = -0.3
• y_com,sys = (m₁y₁ + m₂y₂)/(M) = -0.04

and a couple of the equations I've been trying to work with:

• F_net = Ma_com
• Ma_com = m₁a₁ + m₂a₂

The initial position and CM equations are a good start. You don't have Fnet given, adn you will not be able to find it unless you find the tension in the string. You can do that, but you don't have to. The two masses share a common distance of motion, which means their velocities and accelerations must have the same magnitude. Those magnitudes can be found without actually calculating the tension. You know the directions. From your CM position equations you can come up with the CM velocity and CM acceleration equations.

 

[esoteric deviance]

so what you're saying is:
(change in d₁) = (change in d₂), v₁ = v₂, and a₁ = a₂?

the equations for CM of velocity and acceleration:
v_com = (m₁v₁ + m₂v₂)/(M)
a_com = (m₁a₁ + m₂a₂)/(M)

i don't know what to do...if i set the velocities and accelerations equal to each other, all i end up with is v_com = v and a_com = a.

 

[OlderDan]

so what you're saying is:
(change in d₁) = (change in d₂), v₁ = v₂, and a₁ = a₂?

the equations for CM of velocity and acceleration:
v_com = (m₁v₁ + m₂v₂)/(M)
a_com = (m₁a₁ + m₂a₂)/(M)

i don't know what to do...if i set the velocities and accelerations equal to each other, all i end up with is v_com = v and a_com = a.

The vectors are not equal. The magnitudes of the vectors are equal. The horizontal distance m1 moves is equal to the vertical distance m2 falls