Statics: Exam 3 Review Problem 6, Friction Slipping Tipping
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- Опубліковано 7 лют 2025
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I come over just want to see if you are been well recently. I took the statics classes two years ago and get A+ in my class. You are my best online professors!! Best wish from Berkeley !
I love when you say, "What does the box weigh?" professor :)
also, the calculator voice was lots of fun.
8:43 the whistle
cheers to the best engineering professor out there
DR. Hanson, thank you for another detail explanation of Friction Slipping Tipping problem in mechanics one. These problems continue to be heavily Free Bodied.
Wonderful Statics lessons!! really taking the time to explain the problems along with a good sense of humor :))
I really wish I found your channel earlier. This is really helpful
Thank you very much!
doesn't the center of mass have to be adjusted for the crate and the cart (.92 vs .95) since the cart has mass
I agree- I think there should be two mass forces for the moment equation.
If we want to consider the wheels, the box, and the cart (rectangle) as one body, we must calculate the center of volumetric weight of the body or we deal with the weight of the two bodies as a separate force(If we neglect the weight of the wheels) and not as a resultant correct me guys if i am wrong👍
Good video. I like your down to earth style. Thanks for sharing your knowledge. But shouldn't the cart & the crate be considered separately, because the crate is not centered on the cart?
Please continue the dynamics course! I really need your help.
Thank you u,sir!
im sorry sir, for situation 4, shouldn't it be 359.7? cuz -686.7(.55)/-1.05= 359.7 N
btw thank you sir, it helps a lot and easy to understand
That’s what I got as well
we consider friction in the roller while rotating in example 1 of your, lecture but neglecting it in roller b. why?
I think(but i might ne wrong) that when he says that the roller at point a is locked he meant that friction ONLY applied at point a and not at point b
at 8:50, why isn't the reaction force 'By' included in the friction equation?
Because only wheels at A are locked whereas B is just rolling freely.
I think since they just replace Ay as a normal force "Na" it's not in the ground the friction become opposite to P.
Dr. Hanson, since the crate is not perfectly center on the dolly wouldn't the moment distance for both crate and dolly should be calculate separately for situation 3 & 4?
Yes I was also thinking so, what made me come to the comments. I solved that way and got 195.55N for P when cart slides and 378.4N when cart tips
@Jeff Hanson, Please make a video about C shape columns. And how to apply Secant Formula to find the eccentricity of the given load (given that the eccentricity is on the X-axis) with a numerical problem.
This is actually the question (problem- 1139. Capter-11) from text book of "Strength of Materials (4th edition) by Ferdinand L. Singer and Andrew Pytel.
How did you know there's a break at A?
Can you please update your oyher statics videos? I have exam next week :)) Thank you btw you're the best!!!
Heeyy Professor Hanson, thanks for this video. Has really been helpful. Why doesnt By have a friction component?
Because at the start he states only wheel A is locked, wheel B is rolling
What happen if you can only come up with some scenario but not all of them? I feel like the should a short cut to do it quickly.
hello, how come that the cart and the crate will slide when it locked at point/wheel?
because it's not locked only brakes applied on the wheels
does anyone know why by equals 0 for the last case