Thank you so so so much for doing this video. We have a test on this tomorrow and nobody in our class gets it bc we have a terrible teacher. This made me understand it right away. Thanks so much!
Thanks 06yangji. You can actually set whatever convention that you would like. As long as I have defined that clockwise is positive, then that is what matters. It's all a matter of convention, for if I look at the same exact problem from the other side, I'll get exactly the opposite answer!
@Luuuuuuuuuuuuuuulz You could say that, I suppose. The rod pushes down on the corner, therefor it pushes back on the rod. Any time that you have contact, there will be a normal force. The normal force is perpendicular to the surface, therefore, in the corner, the wall will push perpendicular (to the right) and the floor will push perpendicular (up). Both of these forces are definitely related to Netwon's 3rd law!
helpful... but as 06yanji said, you got your signs wrong on torque because counterclockwise is positive and negative for clockwise. how i remember it is usually due to polar coordinates, the angles are positive if you go counterclockwise and negative if you go clockwise.
Gareth, I need you to clarify a little... I think the answer to your question is a combination of things: The bar is fixed at the bottom (usually the ladder is sliding along the ground), we are ignoring the mass of the bar (often, the ladder does not have negligible weight). I hope this is helpful. If not, please give me an example of exactly what type of ladder problem and I'll try to clarify for you!
If you didn't know the Tension, you wouldn't be able to solve for both (too many variables). However, if you knew the Tension, you can put that value in, and then use a variable for the angle for the shark. Instead of putting sin(30) for the shark, you could use sin(theta). Then you would be able to solve for the missing variable.
@@dandelion44444 Sorry for the late reply, YES I belivee you can do it with the tension in the blue part. You will need to break it up into horizontal and vertical components. Basically the same process, but your unknowns will be in a different location.
This is really helpful but i believe that the torque direction is wrong. Counterclockwise should be positive and clockwise negative. Still a very helpful video. Correct me if I'm wrong.
Austin g no, I was checking emails and I got an email notification saying someone replied to my comment. I was curious so clicked right away. Anyways, this video is great help.
Can I just ask why you use the rod as the lever arm when calculating the torques unlike in a ladder problem the length of the ladder isn't taken as the lever arm?
you have your positive and negative directions wrong. if a force causes a counter clockwise rotation, the torque is a positive term if a force causes a clockwise rotation, the torque is a negative term
plz help me solve this ...angle between two forces of equal magnitudewhen the magnitude of their resultant is equal to the magnitude of either of the forces with diagram
Hi Peter, if the force is perpendicular to the surface of the arm (tangent to the rotation) it may not be necessary to include the angle (as sin (90) = 1).
Hi Naliesha, The torque depends on the angle of the force relative to the bar. If you look at the diagram, the angle of the torque with the bar is 60+20 degrees. That is why it is sin 80. (sin is part of the torque equation)
Thank you so so so much for doing this video. We have a test on this tomorrow and nobody in our class gets it bc we have a terrible teacher. This made me understand it right away. Thanks so much!
Thanks 06yangji. You can actually set whatever convention that you would like. As long as I have defined that clockwise is positive, then that is what matters. It's all a matter of convention, for if I look at the same exact problem from the other side, I'll get exactly the opposite answer!
I think I've got it clear in my head now.
Great video. To solve Fw and Ff, why don't you include the angle 60 degrees when solving for them? Thanks
@Luuuuuuuuuuuuuuulz You could say that, I suppose. The rod pushes down on the corner, therefor it pushes back on the rod. Any time that you have contact, there will be a normal force. The normal force is perpendicular to the surface, therefore, in the corner, the wall will push perpendicular (to the right) and the floor will push perpendicular (up). Both of these forces are definitely related to Netwon's 3rd law!
helpful... but as 06yanji said, you got your signs wrong on torque because counterclockwise is positive and negative for clockwise. how i remember it is usually due to polar coordinates, the angles are positive if you go counterclockwise and negative if you go clockwise.
Thank you!
Gareth, I need you to clarify a little... I think the answer to your question is a combination of things: The bar is fixed at the bottom (usually the ladder is sliding along the ground), we are ignoring the mass of the bar (often, the ladder does not have negligible weight). I hope this is helpful. If not, please give me an example of exactly what type of ladder problem and I'll try to clarify for you!
Dude thanks for this, you are the best! No other youtube video made this more clear! :D
what if the fish is some angle(theta) to the right, so it wont be 90 degrees to down. How can I calculate that 90+theta angle
If you didn't know the Tension, you wouldn't be able to solve for both (too many variables). However, if you knew the Tension, you can put that value in, and then use a variable for the angle for the shark. Instead of putting sin(30) for the shark, you could use sin(theta). Then you would be able to solve for the missing variable.
@@MrBdubsSAS I have tension of blue structure, so I can solve it with its tension too or do I need tension of cable?
@@dandelion44444 Sorry for the late reply, YES I belivee you can do it with the tension in the blue part. You will need to break it up into horizontal and vertical components. Basically the same process, but your unknowns will be in a different location.
This is really helpful but i believe that the torque direction is wrong. Counterclockwise should be positive and clockwise negative. Still a very helpful video. Correct me if I'm wrong.
The convention can be either way.
A blast from the past. 😅Thanks for the correction. Yes, it can be either way.
Wow you must've been watching some youtube because your response was < a minute for a year old comment.
Austin g no, I was checking emails and I got an email notification saying someone replied to my comment. I was curious so clicked right away. Anyways, this video is great help.
Can I just ask why you use the rod as the lever arm when calculating the torques unlike in a ladder problem the length of the ladder isn't taken as the lever arm?
you have your positive and negative directions wrong.
if a force causes a counter clockwise rotation, the torque is a positive term
if a force causes a clockwise rotation, the torque is a negative term
When you get the components of the force, or the radial distance, it doesn't matter, right? You will get the same formula, 4Tsin80???
the x and y forces from the pivot point comes from newtons 3rd law?
plz help me solve this ...angle between two forces of equal magnitudewhen the magnitude of their resultant is equal to the magnitude of either of the forces with diagram
Thank you so much!
Great example.
Do you always have to include the angle when calculating for torque? Because I've seen other examples where this wasn't included.
Hi Peter, if the force is perpendicular to the surface of the arm (tangent to the rotation) it may not be necessary to include the angle (as sin (90) = 1).
can someone please explain when finding the - T tension how did he get sin 80 ?
Hi Naliesha, The torque depends on the angle of the force relative to the bar. If you look at the diagram, the angle of the torque with the bar is 60+20 degrees. That is why it is sin 80. (sin is part of the torque equation)
vry nice i am impressed .
the wording of that problem is really bad for students, great explanation though
I agree. It was taken from an old textbook and students had such a hard time with the wording that I created the video to help out! :)
You sound like Seinfeld.
¿Alguien viendo esto en 2024? :V
Los estudiantes mios todavia la ven cada ano! :)
lol. "my rod"