first i thank god that i found your channel and then i gotta thank you because i got A in my midterms and tomorrow is my finals , and this chapter was my problem. And i just realized that these chapter videos have been uploaded in less than a week , coincidence ? i think not. THANKS SIR!
Shouldn't the distances be based off of the perpendicular Forces applied to points on the lever arm? In other words, shouldn't the distances be lengths on the boom not horizontal distances on the ground? Thanks for the video, it is helping me with one of my homework problems.
Hey Michel, great videos as always. Is there an easier way to find the center of Mass of an object that is on an angle? In this case you were saying the boom was L/3. Can I not just take half the distance of it as if it were horizontal? (After calculating its distance in the horizontal direction). *sorry Im pretty new to this stuff*
@@MichelvanBiezen Thank you so much for your response, Im just wondering because I know the weight of the object, Im just having trouble locating where the center of mass will be on the boom itself, even though its on an angle. Thanks again man, you have saved me hours of stressing!!!
Are you trying to show off how much expensive your shits are. I am watching it in a 6 year old windows laptop using the speaker, and feeling blessed for the contents....
Also, assuming that the boom is uniform, shouldn't the boom's weight vector be applied at a distance of 1/2 L, the location of the boom's center of mass?
Sir, Is there be any tension in the wire containing 1000N .... If yes, then why are we avoiding it when we equate total force in y direction to zero...?
Michel van Biezen Thank you sir, the reason why I asked you this doubt is because I can't see you equating the vertical tension in net force equation..😇😇
Sir, firstly thanks a lot for helping us, and ı cant't imagine how can ı pass the phy 105 without you. Well, ı couldn't catch the difference between cw-ccw. Before this video you took the cw (+), but this video is (-). How can ı decide the direction ?
in the last problem we accounted for the force of the wall on the ladder and in this problem wouldn't the wall exert a force on guy wire ? if the wire werent tied to the wall then wouldnt the system collapse also when we are setting fnety=0 why didnt we account for the tension in the guy wire holding the 1000N
I know your question was from 2 years ago, but in case anyone else had this question, thought it might be useful to put this out there. I think it's because in this problem, the top of the wall is not directly contributing to the beam's tendency to rotate- the tension in the guy wire is. In the ladder problem from previous video, the ladder is directly in contact with the top of the wall, and so the wall exerts a force on it. This time, the wall is in contact with the beam only through the Boom (which is why we are calculating F sub x and F sub y), but not at the top, so no lone-wall force exerted.
F sub x is typically horizontal and F sub y is typically vertical, depending on frame of reference. In most cases, it's like your usual x-y axis directions. They will both contribute components and will result in a Force at some angle phi (as he calculated to be 75 degrees).
first i thank god that i found your channel and then i gotta thank you because i got A in my midterms and tomorrow is my finals , and this chapter was my problem. And i just realized that these chapter videos have been uploaded in less than a week , coincidence ? i think not.
THANKS SIR!
Samer Saeed congratulations for nice grade😆
Excellent video . And love mr bugs bunny in the opera suit 🐰 (right bottom corner)
Shouldn't the distances be based off of the perpendicular Forces applied to points on the lever arm? In other words, shouldn't the distances be lengths on the boom not horizontal distances on the ground? Thanks for the video, it is helping me with one of my homework problems.
thank you so much, you just saved my life
Glad you found our videos! (There are thousands of videos on this channel covering just about every topic in physis). 🙂
A lovely lesson ! Thank you !
You are the best, by the way.
Hey Michel, great videos as always. Is there an easier way to find the center of Mass of an object that is on an angle? In this case you were saying the boom was L/3. Can I not just take half the distance of it as if it were horizontal? (After calculating its distance in the horizontal direction). *sorry Im pretty new to this stuff*
In this example the center of mass of the boom was given. (We can assume that the boom tapers from the base to the end.)
@@MichelvanBiezen Thank you so much for your response, Im just wondering because I know the weight of the object, Im just having trouble locating where the center of mass will be on the boom itself, even though its on an angle. Thanks again man, you have saved me hours of stressing!!!
New camera ? New lighting ? Everything looks great but my left ear is going to hate me after this playlist.
+frnchdazzled I just listened to the video with ear phones and I am getting stereo sound on my computer. What device are you using?
Macbook air, Sennheiser earphones. And i can hear 4-3 little beeps my left ear.
Are you trying to show off how much expensive your shits are. I am watching it in a 6 year old windows laptop using the speaker, and feeling blessed for the contents....
@@debendragurung3033 no he isn't
Hi, I have a question about static equilibrium, I'm not sure how static equilibrium is related to torque concept.
Also, assuming that the boom is uniform, shouldn't the boom's weight vector be applied at a distance of 1/2 L, the location of the boom's center of mass?
If it’s at equilibrium, y is the angle not 60?
To calculate d3 we must use the cos of 30 degrees.
after these videos, are they in the chapter static equilibirum or not ?
We have a lot of these types of videos in the static equilibrium playlist as well.
You are awesome!!! Thank you!
Happy to help!
Thank you!
You're welcome!
Sir, Is there be any tension in the wire containing 1000N .... If yes, then why are we avoiding it when we equate total force in y direction to zero...?
The 1000 N appears in both the force in the y-direction equation and the torque equation. (it is not being ignored)
Michel van Biezen Thank you sir, the reason why I asked you this doubt is because I can't see you equating the vertical tension in net force equation..😇😇
Sir, firstly thanks a lot for helping us, and ı cant't imagine how can ı pass the phy 105 without you. Well, ı couldn't catch the difference between cw-ccw. Before this video you took the cw (+), but this video is (-). How can ı decide the direction ?
As a vector quantity, CW (is negative) and CCW (is positive). If you are only finding the magnitude then it doesn't matter what direction you choose.
in the last problem we accounted for the force of the wall on the ladder and in this problem wouldn't the wall exert a force on guy wire ? if the wire werent tied to the wall then wouldnt the system collapse also when we are setting fnety=0 why didnt we account for the tension in the guy wire holding the 1000N
Yes, the force on the wire is the T (tension) which is part of the equation.
I know your question was from 2 years ago, but in case anyone else had this question, thought it might be useful to put this out there.
I think it's because in this problem, the top of the wall is not directly contributing to the beam's tendency to rotate- the tension in the guy wire is. In the ladder problem from previous video, the ladder is directly in contact with the top of the wall, and so the wall exerts a force on it. This time, the wall is in contact with the beam only through the Boom (which is why we are calculating F sub x and F sub y), but not at the top, so no lone-wall force exerted.
Thanks a loooooooooooooot 🙋
Tanks =)
You are welcome. 🙂
Hello sir
Can You explain me about the direction of F sub x And F sub y?
F sub x is typically horizontal and F sub y is typically vertical, depending on frame of reference. In most cases, it's like your usual x-y axis directions. They will both contribute components and will result in a Force at some angle phi (as he calculated to be 75 degrees).