| Problem statement Solution video |
DISCUSSION THREAD
NOTE: The system is released from rest in a vertical plane.
DISCUSSION
Recall the four-step plan for solving kinetics problems.
Step 1: FBDs
Here it is recommended (as for all problems using the Newton/Euler approach) to draw individual FBDs for the system: one each for block A and block B, and for the pulley.
Step 2: Kinetics (Newton/Euler)
Write down the Euler equation for the pulley, along with the Newton equations for A and B in the vertical direction. With that, you have three equations and five unknowns (TA, TB, aA, aB and αpulley). Please note that the tensions TA and TB are NOT equal to the weights of blocks A and B.
Recall from ME 270 that friction opposes (impending) motion. For this problem, you need to decide in which direction (CW or CCW) the pulley will be rotating. To do this, consider first the friction-free case. Once the direction of rotation is determined from that, draw the pulley FBD with the friction torque opposing that motion.
Step 3: Kinematics
It is recommended that you write down the acceleration of points C and E on the pulley (see figure above) in terms of the acceleration of the pulley center O. Then, relate the accelerations of A and B to the accelerations of C and E, respectively.
Step 4: Solve
At this point, you have five equations and five unknowns. Solve!
Any questions?? Please ask/answer questions regarding this homework problem through the “Leave a Comment” link above.
Are we okay to find the total moment and inertia in the pulley and blocks and solve for angular acceleration right away? Is this shortcut valid for future problems or is this a special case?
I would recommend against it. Your “shortcut” here will not work for many of the other problems that you will see of this type in the course.
I had the same question. I was also wondering how the problem would be different (if it would be) if the cable was like extendable instead of inextensible/having a fixed length? Would we have to add in more kinematic equations then?
When assuming the initial rotation direction, can we still make counter clockwise positive even if it’s initially gonna rotate clockwise when there’s no friction?
Setting up sign conventions is not the same thing as assuming that the system moves one way or the other. You need to set up your sign conventions and stick with them in a way that they are consistent from one FBD to another and from one kinematics equation to another.
To find which way the pulley rotates, could we calculate the torques acting on the pulley from A and B, and whichever torque value is greater would be the rotation of the pulley? In this sense, then the friction force would act opposite to this?
The general rule of thumb I used to define the signs in my problem was to look at the impending motion of the pulley to see which way the accelerations of C and E faced. Using this information I set my positive y_A and y_B in those respective directions. Then for the moment I followed general right hand rule for sign convention.
**the sign convention of the rotation should follow the sign convention directions for the blocks as well. Therefore in this problem if choosing the impending motion of the blocks as their positive directions the CW rotation should be positive.
for the Pulley’s FBD should the reaction forces at O be included?