Great video! Would you mind explaining how to figure out the tipping point for a pedestal table. I’m having trouble deciding the amount of allowable overhang past the table base. Thanks.
Huh, I must be misunderstanding how this is all done. Of course, I only did up to early college calc a couple decades ago, so my skills are terrible. I'm just interested in the math for a problem. Why is the axis, which I would assume would be the point where the beam would rotate around, be at N0? You said you wanted to solve for X, the point where the man would be standing on the overhang, where it would lift the beam off N0. So, wouldn't that make the axis point be over N1, since the beam would be rotating over that post? Or, is the axis the area where the range of motion will be? Sorry, just confused by the lingo. Maybe I'm mixing up axis for axle.
When objects are in equilibrium (net force = 0 and net torque = 0) that means that you can pick ANY axis of rotation you want. Obviously, if the object has not started rotating, it is not rotating about any axis. So you can absolutely pick a different axis of rotation and if you're careful with your math, you should end up with the same result. Cheers, Dr. A
It's so helpful....but I don't understand smth: Why we don't have a normal force acting on the place where the man is on the beam??? Why you didn't mention it....I sow another problem where also they didn't take in consideration that force, but I don't understand why since the man is over the surface???? There are n1 and n0 force but where is the normal force acting on the man???
The second to last equation has the torque: force * lever arm. The very last equation is setting the torque equation equal to zero and solving for x. Try it yourself on paper. Cheers, Dr. A
Wonderful lecturer that makes physics fun and understandable !!!
Classic static problem. Thanks for sharing.
+Yungmo01
Yungmo,
Yes this is definitely one of the classic problems. I'm hoping to hit them all soon.
Cheers,
Dr. A
Wait a second... is this guy writing backwards?
Nope. Secret here: www.learning.glass
Cheers,
Dr. A
Great video! Would you mind explaining how to figure out the tipping point for a pedestal table. I’m having trouble deciding the amount of allowable overhang past the table base. Thanks.
Thank you so much!
Muhammad Rashid,
You're very welcome. Glad you're enjoying the videos.
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A
Thanks. You really helped me out.
+scratch12367
You are very welcome. That's the goal is to help people out.
Cheers,
Dr. A
god bless you young man 🙏
You rule.
Thanks. I'm enjoying it.
Cheers,
Dr. A
thanks sir
Huh, I must be misunderstanding how this is all done. Of course, I only did up to early college calc a couple decades ago, so my skills are terrible. I'm just interested in the math for a problem. Why is the axis, which I would assume would be the point where the beam would rotate around, be at N0? You said you wanted to solve for X, the point where the man would be standing on the overhang, where it would lift the beam off N0. So, wouldn't that make the axis point be over N1, since the beam would be rotating over that post? Or, is the axis the area where the range of motion will be? Sorry, just confused by the lingo. Maybe I'm mixing up axis for axle.
When objects are in equilibrium (net force = 0 and net torque = 0) that means that you can pick ANY axis of rotation you want. Obviously, if the object has not started rotating, it is not rotating about any axis. So you can absolutely pick a different axis of rotation and if you're careful with your math, you should end up with the same result.
Cheers,
Dr. A
@@yoprofmatt oh, okay. Thank you.
Could you make a video about your set up? What are you using to project the transparent board? Thanks!
Check our webinars at www.learning.glass
Cheers,
Dr. A
It's so helpful....but I don't understand smth: Why we don't have a normal force acting on the place where the man is on the beam??? Why you didn't mention it....I sow another problem where also they didn't take in consideration that force, but I don't understand why since the man is over the surface???? There are n1 and n0 force but where is the normal force acting on the man???
The normal force is applicable in the free body diagram of the man - not the FBD of the beam - as it acts on the man, not the beam.
thx
great video but bruh you lost me on looking at you in the background
This was as clear as day to me right up until the last equation. How did you derive that?
The second to last equation has the torque: force * lever arm.
The very last equation is setting the torque equation equal to zero and solving for x. Try it yourself on paper.
Cheers,
Dr. A