I am having trouble finding the given answer and I am fairly certain I'm making a conceptual mistake.
I started with the summation of forces but that didn't yield anything interesting so I went to the Euler equation, about point a.
W = weight
Sum Ma = W beam (distance) + W winch (distance) + Oy (distance) + Force winch( distance)
Ma = -1/3 m L^2(alpha) - W winch d + Oy d + torque*(1/rwinch)*ra
alpha = (acceleration)/radius (a to COM) = 32.2/ 8
Plugging in and solving I get Oy = 1677.0625 lb
I think I am going wrong with either the I beam weight or the winch force.
7 comments:
Here are a few comments on your solution that come to mind:
* Carefully consider your FBD. If you make your FBD by cutting through the cable and the right hand support at O (as shown in the FBD in the original post), you will have five forces: the cable tension T, W_beam, W_winch, O_x and O_y.
* The tension force is the same as the "winch force" you show above: torque/r_winch.
* Note that the acceleration of A is NOT zero. Therefore, you can NOT use the "short form" of Euler's equation if you choose point A for the summation of moments. My recommendation would be to use Euler's equation about the fixed point O rather than about the moving point A.
* Finding the mass moment of inertia (MMI) is a little tricky. You have just included the MMI of the beam about O. You need to add on the MMI of the winch about point O. For a point mass, the MMI is m*d^2 where d is the distance from the point mass to point O.
Let us know if you have questions on this and/or if this does not get you to the correct solution.
Thanks for the help, unfortunately it opened a new can of worms.. the answer I am now getting is much further than the first answer (5798.5).
Solving for Mo
Mo= mwinch/g(r^2)alpha
+1/3(mbar/g)(r2^2)alpha
+T(d)
since the bar is "held in horrizontal possition" Mo=0. After this I solved for alpha (12.508) and multiplied it by the distance it is from O to find the values for the sum of y eqn.
-Oy+mwinch(awinch)+mbar(abar)- T =0
solving for Oy yielded Oy = 5798lb
the bar force was very high (6,220.95 lb)and it is probably the source of the problem. But I checked the components and cannot find the error. alpha)*r= a
a*m=F
Let me say first that the author's have used a poor choice of wording when they say that the "beam is held in a horizontal position". This appears to say that the beam is in static equilibrium when it is not (it has an angular acceleration). What is intended here is for you to determine the vertical component of the reaction force at O for the given value of motor torque. Do NOT set M_O equal to zero.
Here is a sketch of what you should have for your Euler equation:
sum M_O = I_O*alpha
where
sum M_O = 4*W_winch + 8*W_beam - 16*T
I_O = (1/3)*m_beam*16^2 + m_winch*(4^2+2^2)
Use this equation to find alpha.
Then use Newton in the y_direction to find Oy:
sum Fy = M*a_Gy
where
sum Fy = T - W_beam - W_winch +Oy
M*a_Gy = m_beam*(-8*alpha) + m_winch*(-4*alpha)
These equations are not all that far from your equations; however, they follow from a straight-forward application of the Newton/Euler equations.
One source of problem with your solution, I think, is that you are confusing mass with weight. In your mass moment of inertia you are dividing mass by g. You need to divide weight by g to get mass. In your Newton equation you are not dividing weight by g.
I hope that this has not been too confusing and has helped. If it has not helped, let us know.
I am getting an answer of 1419 lbs, the book's answer is 1469 lbs.
These are the values I calculated:
T = 1200 lbs
M_0 = -1200 ft-lbs
I_0 = 5615.95
alpha = -.2137
Am I making a mistake somewhere?
I got -1700 for the sum of the moments about O (and I also got the book answer). Don't forget to add the radius of the pulley on the end of the bar, and convert to feet, since the tension force acts there.
Thank you! I was having the same problem but your suggestion fixed it.
-John
I'm not sure how we determined the radial distances corresponding to a_Gy to be negative.
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