I know the bar is somehow going to swing to the right and go down, but are pt. A and pt. G constrained to a particular type of motion (i.e. horizontal or vertical)? Anyone have any idea on what I'm missing? Your help is appreciated.
Apr 15, 2008
Sample Exam Problem 24
I know the bar is somehow going to swing to the right and go down, but are pt. A and pt. G constrained to a particular type of motion (i.e. horizontal or vertical)? Anyone have any idea on what I'm missing? Your help is appreciated.
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You are on the right track on your thinking.
Note that the system is at rest (all omega's are zero on release).
* Therefore, when you write down the kinematics equation for AD, you will see that the acceleration of A is perpendicular to line AD (horizontal in this case). Importantly, you have only one scalar unknown for a_A.
* When you write down the kinematics equation for GE, you will see that the acceleration of G is perpendicular to line GE. Importantly, you have only one scalar unknown for a_G.
* When you write down the kinematics equation for link AG, you will have two scalar equations. This added to your three Newton-Euler equations gives you five equations. And you will have five unknowns: T_AD, T_GE, a_G, a_A and alpha.
I understand the logic now. Thanks a lot professor. I still have a doubt on what role a_A plays on this problem. The Newton-Euler equations I get don't reference a_A in any way. For the kinematics for the bar I end up with:
a_gx = a_a
a_gy = alpha*(L/2)
and using trig I get
a_gx = a_g*sin(theta)
a_gy = a_g*cos(theta)
Do I have to account for the signs when I substitute a_gx and a_gy into the Newton-Euler eqns or ignore that since signs are already accounted for in the sum of forces?
I hopes this doesn't sound too confusing. Thanks.
The acceleration of a (a_A) is just another unknown in the problem. You can solve for that (if you want) along with the other unknowns in the problem.
Your kinematics are correct except for the signs on the components of a_G. Since a_G is perpendicular to GE, you can write EITHER:
a_Gx = -a_G*sin(theta)
a_Gy = a_G*cos(theta)
OR
a_Gx = a_G*sin(theta)
a_Gy = -a_G*cos(theta)
That is, a_Gx and a_Gy must have OPPOSITE signs in order for it to be perpendicular to GE (make a sketch to see this).
Does this help?
Yes I did understand that. My a_G ends up being perpendicular to GE and pointing down:
a_Gx = a_G*sin(theta)
a_Gy = -a_G*cos(theta)
But my question is do I substitute these expressions with their respective signs into the sum of forces: ex. the sum of forces in y is T_ad + Teg*sin(theta) - Mg = Ma_Gy.
In the Ma_Gy part would I substitute a_Gy as -a_G*cos(theta) or do I omit the sign since the forces already take the direction into consideration??
You should NOT omit the signs as it is important to keep the relative signs correct for the correct direction. Putting both a_Gx and a_Gy into Newton with positive signs would be incorrect as a_G would point the wrong way.
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