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Tuesday, September 25, 2007

Homework Hint: Problem 7/24

Suggestions:
  • Attach observer to boom of crane.
  • Use point O as the reference point for the velocity and acceleration equations. Note that this is a legitimate point to use since it is on the same rigid body as the observer.
  • Note that (v_P/O)_rel = (a_P/O)_rel = 0 for this choice of observer.

13 comments:

kgjohnso said...

Is there anyway you can post the answers to this problem so I can check my answer?

Deborah said...

which way does Beta_dot rotate because i do not know if a crane moves up and down or around the z axis?

kgjohnso said...

the crane is being lowered @ a rate of beta dot. Looking at the picture the boom rotates CW

Deborah said...

so is beta_dot in the x direction

kgjohnso said...

The way i put my axis beta_dot was in the -x

ccb056 said...

kgjohnso:

I got the velocity to be:
2.08i - .12j - 2.51k

and acceleration to be:
.44k - .41i

kgjohnso said...

I got very similar Vel. but i had 1.2 not .12. And for the accel. i guess i need to go back thru my work

ccb056 said...
This comment has been removed by the author.
ccb056 said...

I'm not sure I did the problem correctly....

What did you use for Vo and Ao?
I set these to 0, im not sure that is correct...

Deborah said...

i set vo and ao to zero also

CMK said...

As you compare answers, be sure that you are using a common set of coordinate axes. For example, you could put x horizontal, y vertical and z out of the page. Or, alternately, you could align x with the boom, y perpendicular to the boom (up and to the left) and z out of the page. Or, ...

Your final vector answers will depend on your choice of coordinate axes. Regardless of which coordinates that you use, however, the magnitudes should be the same. For the sake of comparison, I found the following magnitudes:

v_P = 3.48 m/sec
a_P = 1.10 m/sec^2

kgjohnso said...

thanks ck

ccb056 said...

Well, I've learned to trust my calculator when it comes to these long cross products ....