Complicated pulley system on an incline?

[Hema]

Homework Statement

Under the action of force P the constant acceleration of B is 6 m/s^2 up the incline as in figure. For the instant when B's velocity is 3 m/s up incline, what is the velocity of point C? How do I solve this using constraints?
[/B]

Homework Equations

For a pulley system ΣT.v = 0 since internal work done adds up to zero
T*2-2T*3+TVc = 0 (given in my book) which gives Vc = 4 m/s.

The Attempt at a Solution

It had been previously proven that Va = 2 m/s at the instant mentioned. I decided that the equation T*2-2T*3+TVc = 0 was for the points A, C and D. Was I right? Am I allowed to take points C and D for the equation? If I am, why is it so? If the equation is not for A, C and D then for which points is it? The solution had previously used A and B for the constraint equation to find A's velocity, using the equation -2T*3 + 3TVa = 0.

 

[Charles Link]

I see a simple way of doing this last part to check your answer. Look a the top section of rope. It needs to lengthen according to $v_B-v_A$. Point C will move forward at this same speed relative to B, and B is moving. This is at least one way to get the answer. ( Once you have the result that $v_A=2 \, m/sec$, the part to find the speed of the point C on the rope is fairly simple). $\\$ Edit: It can be helpful in problems like this to consider the rope as a fixed length $L$ and constraint equations can then be written by describing the rope as $L=L_1+L_2+L_3$, and then derivatives can be taken. e.g. $L_1=x_B-x_A$, so that $\frac{dL_1}{dt}=v_B-v_A$. Also $\frac{dL}{dt}=0$.

 

[haruspex]

T*2-2T*3+TVc = 0

I decode that as T.V_a-2T.V_b+T.V_c. That appears to come from considering the section of cord from A (upper), passing over and including the pulley at B, and back to C.

Note that this way of looking at it requires figuring out that the tension in the cord tying the upper pulley to B is 2T. Alternatively, don't include that pulley in the subsystem, and define the subsystem as the combination of the straight section of cord above the pulley and the straight section below the pulley back to C.

As with any mechanics problem, you can specify whatever subsystem you like and consider the external forces on it.

Another option is to choose the cord from C down and around the lower pulley to D. This gives T.V_c=2T.V_a.

 

[Hema]

I decode that as T.V_a-2T.V_b+T.V_c. That appears to come from considering the section of cord from A (upper), passing over and including the pulley at B, and back to C.

Note that this way of looking at it requires figuring out that the tension in the cord tying the upper pulley to B is 2T. Alternatively, don't include that pulley in the subsystem, and define the subsystem as the combination of the straight section of cord above the pulley and the straight section below the pulley back to C.

As with any mechanics problem, you can specify whatever subsystem you like and consider the external forces on it.

Another option is to choose the cord from C down and around the lower pulley to D. This gives T.V_c=2T.V_a.

Thanks! Would you please explain to me how we can define the subsystem without the pulley and only the straight sections of cord above and below? Won't that just give the equation T.Va + T.Vc = 0?

 

[haruspex]

Thanks! Would you please explain to me how we can define the subsystem without the pulley and only the straight sections of cord above and below? Won't that just give the equation T.Va + T.Vc = 0?

No, the tension acts in the same direction on that subsystem at A and C. We need to oppose those with the tensions exerted by the pulley at B. The trick is that the sum of the two velocities there is 2V_b.