| Problem statement Solution video |
DISCUSSION THREAD
NOTE: Consider all surfaces to be smooth. Leave g as part of your answer.
Discussion and hints:
Let's first take a look at the motion of the mechanism, as shown in the simulation results below.
Recall the following four-step plan outline in the lecture book and discussed in lecture:
Step 1: FBD
Draw a free body diagram for the entire system of A+B+bar. Which, if any, forces do non-conservative work on this system? Can you justify this from the FBD?
Step 2: Kinetics (work/energy equation)
Write down the work energy equation for the system of A+B+bar. The KE will be the sum of the KEs of A and B. The PE will be the sum of the PEs for A and B.
Step 3: Kinematics
Look for the instant center at Position 2 to simplify your kinematics for that position.
Step 4: Solve
Solve your work/energy equation for the speed of block A.
Any questions??
Just to clarify, since it differs from the image in the discussion post, A and B both have mass of m, right?
I would go off of what is in the problem statement and assume they are both mass m.
Yes, please use the fact that both blocks have the same mass "m". I have corrected the figure above. Thanks for bringing this to our attention.
It says to leave our answer in terms of at most m, theta and L. I also have g in my answer so should I plug in 9.81 or 32.2 for this or should there not be a g? Also, the animation and image on the website are quite different than the one on the problem pdf, so should I assume that A can move below the position of B as shown in the animation or that it stops once it reaches the bottom of the slot (same height as B) as in the pdf.
What happens after the instant of interest has no bearing on what happens before the instant of interest. The motion shown in the animation is completely relevant to the question that is asked of you here.
Should we have g in my answer? I also have g in my final answer.
yes, g may be in your final answer
If I use a system with both blocks and the bar, do I need to account for the internal forces between the bar and each block, which are not perpendicular to the motion of either until position 2?
You should not need to use the internal forces if your system is the bar and both blocks.
I’m not understanding physically or conceptually how the velocity of block B will be 0 once block A reaches the ground. To me it seems like block A will slowly gain speed until it reaches its highest speed the instant it reaches the ground therefore block B will exhibit the same behavior.
This was a fascinating point to ponder for me as well. It's not immediately intuitive, but I think that at some point, the force from the bar switches from compression (pushing against the blocks) to tension (pulling away from the blocks). At the start, it's in compression from the motion of A downward, but at some point before A reaches the bottom, it changes to tension and pulls against B's motion until v_B = 0. It's helpful to use the instant centers approach to find the instant center of the bar when A is at the bottom, too, since the instant center has zero speed at that moment.
If there were no ground, and A kept going downwards, the point at which A and B are level is the point at which block B would be changing directions instantaneously. The acceleration of B is nonzero, but since it is changing directions, the velocity is zero. This is similar to the peak of the path of a projectile, where at the top of the peak, the vertical velocity of the projectile is zero. A mathematical way to understand this would be to use the Pythagorean theorem to write an equation for the positions of A and B in terms of the bar L, then derive it, similar to the tension problems covered previously in class.
For the potential can we use two datums, one for block A and one for B like in example 4.B.6 as that would make V1 =0 and that would make the calculations a lot easier
In theory yes, you can use separate datums for different blocks but I don't see the point at B does not move vertically.
It's safe to assume the walls are smooth right?
Yes, all surfaces are smooth.
Since, we don't consider the force of the bar on A and B, and the normal force doesn't do work, is it correct that there are no non-conservative forces?
Yes. The only force that could be considered an non-conservative force is the tension from the bar. However, this force cancels out, as it is equal and opposite between the two blocks.
The concept of the instant center is very important in this problem. If you find that the distance from an instant center is zero, the velocity of that point is zero.
Are we allowed to keep g in our answer or have I done something wrong?
The instantaneous center method is valid to use. You can also rigid body kinematics to relate the velocity of B and A (the equation that includes angular velocity), knowing that the velocity of B moves in the -i direction, and the velocity of A moves in the -j direction based on the simulation provided.
For everyone: you may leave the constant g as part of this answer
If I use IC to solve, can I solve for omega using the V_a and V_b equations? I am confused about this because I learned tht V_b =0.
Note that your work/energy equation resulting from Step 2-Kinetics has two unknown: v_A2 and v_B2. You say above that you learned that v_B2 = 0 (I assume that you found this from kinematics such as instant centers.) This leaves your work/energy equation with only one unknown. Solve that equation for v_A2.