When you sum moments about a point other than the mass center (in this case b) you need this equation: sum moments about b = Moment of inertia about g (mass center) * alpha + vector between g and b cross mass * acceleration of mass center.
Is it correct to say that the acceleration of the mass center of the system (G) is lineair to the acceleration of point C? At 17:41 you state that the total sum of forces (of the full system) in the y-direction is equal to the mass and acceleration of just the spool…?
Only the spool is accelerating and we are treating the spool and cart as a single system, so the total sum of forces (of the full system) in the y-direction is equal to the mass and acceleration of just the spool.
In 12.40: took moment about wheel B and got NA = 470, NB,570. Any suggestions? If we were to take the moment about any point we should get the same answer ..from statics course
I took the sum of the moments about point B at 13:04 and got 470.4 N as the answer. Any suggestions as to why I got a different answer? I did that to eliminate an unknown.
Does anyone know why in 14:37, the professor was able to change the summation of moments about b to rc with respect to b multiplied by mass of the spool?
Thank you, but you said earlier in the video that if a rigid body undergoes only translation, all of the particles of the body have the same acceleration, so a sub G = a sub C, so if that is true why would you change r of G with respect to B to r of C with respect to B. That would give you a different answer
The spool's mass center is at C, not at G. So when summing moments about B, I use the vector from B to C and the acceleration of the spool's mass center. Remember, the equations require the location of the mass center of the body, not just any point on the body.
@@ashlynnundlall Since only the spool is moving I rewrote the equation as sum Moments about B = Rc/b cross mass * acceleration of the spool. Rc/b is to the left and up.
I can not hear this.
In 14:00 while calculating the moment about point B, why you took the moment of inertia of G mass center not C ?
When you sum moments about a point other than the mass center (in this case b) you need this equation: sum moments about b = Moment of inertia about g (mass center) * alpha + vector between g and b cross mass * acceleration of mass center.
Is it correct to say that the acceleration of the mass center of the system (G) is lineair to the acceleration of point C? At 17:41 you state that the total sum of forces (of the full system) in the y-direction is equal to the mass and acceleration of just the spool…?
Only the spool is accelerating and we are treating the spool and cart as a single system, so the total sum of forces (of the full system) in the y-direction is equal to the mass and acceleration of just the spool.
@@ColinSelleck Alright thank you sir! Got my test tomorrow and feeling a lot better about it because of your videos, thanks again!
@@roanrhemrev1591 You're welcome and good luck!
For example 1, 9:04 why isn't the frictional force at the wheels accounted for?
We were told to neglect the mass of the wheels, which is code for ignore the friction of the wheels.
Okay. Thank you very much.
In 12.40: took moment about wheel B and got NA = 470, NB,570. Any suggestions? If we were to take the moment about any point we should get the same answer ..from statics course
I made a mistake in the video Na is 470 and Nb is 571.
Try to be louder next tym plizz
I took the sum of the moments about point B at 13:04 and got 470.4 N as the answer. Any suggestions as to why I got a different answer? I did that to eliminate an unknown.
Hard to say what the error was but it should give you the same answer.
Does anyone know why in 14:37, the professor was able to change the summation of moments about b to rc with respect to b multiplied by mass of the spool?
That is because only the spool is accelerating in this composite body.
Thank you, but you said earlier in the video that if a rigid body undergoes only translation, all of the particles of the body have the same acceleration, so a sub G = a sub C, so if that is true why would you change r of G with respect to B to r of C with respect to B. That would give you a different answer
The spool's mass center is at C, not at G. So when summing moments about B, I use the vector from B to C and the acceleration of the spool's mass center. Remember, the equations require the location of the mass center of the body, not just any point on the body.
at 8:14, isnt Mb supposed to be minus instead of plus in between as per the textbook?
Yes - it should be ema(t) - hma(n)
in the third example in minute 13:36, why are NA & NB multiplied by 2?
Because there are four wheels and due to symmetry, the front wheels have equal loading as do the rear wheels.
They say there are four wheels so of course two at the back and two at the front do you take into account each wheel.
The Na and Nb forces would be the normal force at each respective wheel.
You should add this to the playlist of Ch: 17 :)
Just added it. Thanks for the tip!
Why is the vector Rb/g not Rg/b as it goes down and to the left 16:33
The equation always uses the vector from the point you are summing moments about to the mass center: Rg/b.
@@ColinSelleck But is that not up and to the left then sir in this video. Not down and to the right
@@ashlynnundlall Since only the spool is moving I rewrote the equation as sum Moments about B = Rc/b cross mass * acceleration of the spool. Rc/b is to the left and up.
16:32 shouldn't the i component be -0.5 because it is r of G with respect to B
Sorry, just saw this. Since only the spool is moving, I rewrote the equation in terms of r of C with respect to B.