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Jul 24, 2008

Problem 8/61


Steps:
  1. FBD of block: pay careful attention to both the DIRECTION and MAGNITUDES of the spring forces acting on the block that are shown in the FBD above.
  2. Newton/Euler: sum F_x = 2*(k/2)*(x_B - x) = m*x_dot_dot
  3. Kinematics: none needed
  4. Solve: Note that the "base excitation" here provides a prescribed motion of the box as a function of time. In your EOM in 2., this is an "inhomogeneous" term that is harmonic with a frequency of omega. Solve for the steady-state response for x(t) = X(omega)*sin(omega*t). Make a sketch of X vs. omega. From this determine the ranges of omega over which the absolute value of X is less than 2*b.
CORRECTION on Step 4. above. We are asked to make the amplitude of the motion of the block RELATIVE to the box less than 2*b. The motion of the block relative to the box is given by z(t) = x(t) - x_B(t), not x(t). Use your result for x(t) in Step 4. to find z(t) = (X - b)*sin(omega*t). Make a sketch of X - b as a function of omega, and from this, find the ranges of omega over which the magnitude of (X - b) is less than 2*b. Sorry for any confusion created by my above hint. Again, I did not read the question very closely.

Let us know if you have questions on this.

8 comments:

paul m said...

i am confused as to why F 0 is not equal to 0 isnt sin(0)= 0?

Engineering Help Page said...

You are right about sin being equal to zero but I think
xb=b sin (w)t = Fo sin w(t)

b = Fo

I am confused about how to do this.

I started out looking at this formula.

X = ( b / (1-(w/wn)^2))

and tried to plug in 2b = X and found sqrt(1/2).... and no second value. How am i going about this wrong?

bkorty said...

How do we know that z(t)=x(t)-x_B(t). Are we assuming x(t) is positive because of the statement of the problem asking for m "relative to the cart"?

Another way to put it is: why is it not x_B(t)-x(t) instead?

Tom Gentry said...

I am having trouble understanding how to relate the compression of the springs to the motion of the box. I know that the box is moving with x = b*sin(w*t) and the force on Ma (F) is F = (Ma)*a = k*(compression of spring). I am unsure where to go from here.

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

bkorty,

And you can assume x_B(t)-x(t), you just have to flip the arrow on the FBD.

CMK said...

--to bkorty--

The question asks for the motion of the block relative to the box. If you are an observer on the box, the motion that you would see of the block is:

z_block/box = x_block - x_box = x - x_B

If you write x_B - x, you are describing the motion of the box relative to the block.

Tom Gentry said...

So with x_block/box = x_block - x_box

Fx = m*a_block = m*x_block_dot_dot = k*x_block

x_block/box_dot_dot = x_block_dot_dot - x_box_dot_dot so

m*(x_block_dot_dot + x_box_dot_dot) = k*(x_block - x_box)
We know x_block = bsin(wt) so x_block_dot_dot = -b*w^2*sin(wt) leaves us with:

m*(x_block_dot_dot + b*w^2sin(wt)) = k*(x_block - bsin(wt))

x_block_dot_dot + k/m*x_block = (k/m*b-b^2*w^2)*sin(wt)

Is this correct?