| Problem statement Solution video |
DISCUSSION THREAD
DISCUSSION
Using the four-step plan:
STEP 1: Free body diagram (FBD) – Draw an FBD of bar AB.
STEP 2: Kinetics – Write down the Newton/Euler equations for the bar based on your FBD above. For the “short form” of the Euler equation, please note that you are constrained to using a moment about the center of mass G since there are no fixed points on AB; that is, you should use ΣMG = IG α.
STEP 3: Kinematics – With the inextensible cable being taut, all points on the rigid body AB have the same acceleration, and the angular acceleration of AB is zero: α = 0.
STEP 4: Solve – Use the equations from STEPS 2 and 3 to solve for the tension in the cable and the reaction at on the bar at A.
Any questions?? Please ask/answer questions regarding this homework problem through the “Leave a Comment” link above.
When solving this, should we treat the bar as undergoing pure translation with the base, or do we need to consider rotation about point A when writing the equations of motion?
The problem statement says that the cable prevents rotation of the bar. Therefore, there is no rotation of the bar (alpha = 0).
If your question is in regard to which point that you should use for the moment (Euler) equation, then the answer is that you should NOT use point A. Point A is neither a fixed point nor the center of mass. Because of this, you cannot use the “short form” of Euler’s equation with point A. You may, however, with the center of mass of the bar.
When summing the forces for A in the j direction (if you establish j is vertical), should you account for gravity? The problem does not mention it but i take it that such an assumption is expected by this point in the course…
I assume this is the case since there is no mention of being in the horizontal plane
Either with or without weight is fine in your solution.
In the moment equation about A, why does the weight produce zero moment even though it acts downward on the bar?
There are couple ways to explain this, although they are all based on the same thing. Both are based on what you learned in ME 270.
One argument is to recall that the moment about a point A due to a force is the force times the “moment arm” of the force, where the moment arm is the perpendicular distance from the line of action of the force to point A. Since the line of action of the weight force passes through point A, the moment arm is zero, and therefore, the moment about A is zero.
Another argument is based on the definition of the moment about a point due to a force F:
M_A = r_G/A x F
For this problem, the weight force vector is parallel to the vector r_G/A. Since the cross product of two parallel vectors is zero, then the moment about A due to the weight is zero.
I have a question for you. You are asking here why the weight creates a zero moment about point A. How did you know that it created a zero moment in the first place?