WELCOME TO THE ME 274 BLOG

Here you can add discussion posts as well as add comments to existing posts. Links are also provided to solution videos and other pages of the course website for your easy access.

__________________________________________

THE BLOG

TO ADD A POST, click here.

Saturday, September 8, 2007

Homework Hint: Problem 3/161


A couple suggestions:
  • Treat the sphere (A), block (B) and two links as a single SYSTEM. With this choice, the kinetic energy is the sum of the KE due to the sphere and the block. Also with this choice, only forces external to the system need to be considered when determining the external work done on the system.
  • Use the INSTANT CENTER method to determine the velocity of B at the final position. In particular, locate the IC for link BC when theta = 180 degrees -- what does the location of this IC imply about the speed of B?
  • Be careful on the signs of the potential energies for A and B. In doing so, choose datum positions for A and B. If the particle moves above the datum for the particle, then the potential energy change is positive. If the particle moves below the datum for the particle, then the potential energy change is negative. As you can see, the change in gravitational potential energy for A and B will have opposite signs.

4 comments:

Matt.K said...

Because this problem does not have any forces or friction, and the normal force is perp. to the direction of motion of the the 4Kg block (B)I do not see where there would be any external forces. Does this make sense? I feel like I might be trying to over simplify it.

CMK said...

No, your are not oversimplifying; your conclusion is correct.

Treating A and B (along with interconnecting rods) as part of the system, the only external forces (other than weights) acting on the system act either at a fixed point (O) or perpendicular to the path of the point at which they act (B). Weight forces are conservative and therefore, can be included in potential energy rather than work.

Therefore, no external work is done on the system by non-conservative forces.

Ali Azhar said...

what other method is there to find velocity of 4kg carriage(B)?

CMK said...

If you do not want to use the IC method to show that v_B = 0 for theta = 180 degrees, you could use the standard vector approach. That is, write a velocity equation for point C using link BD and a velocity equation for point C using link OC. Equate these and break up into x- and y-components. From these two equations you will see that v_B = 0 for theta = 180 degrees.

Does that make sense?