I just started taking a mechanics of materials CE class at the polytechnic school in California.. I am using this video for statics review to help prepare me. Thanks!
Hi Mark, great content. About the last question. I did force balance in x direction. My argue is that if the sum is ZERO then the body will stay at rest. If Fx +Ff > P then body will move down. Otherwise if Fx + Ff < P then body will move up. The trick here is on what direction friction force is acting (up or down). Appreciate feedbacks.
The moment in question 2 should be 866 kN-m and not 946 kN-m. I plugged in the numbers half a dozen times exactly how you have them written and get the same answer. The final distance for a force of 226.6 kN should then be 3.82 m.
@Mark Mattson : Regarding the last problem: " Realizing that full friction of 94.75 N does not get generated /mobilized until it is demanded by net forces is the key" In this problem only 81.94 N or slightly more (101.56-19.62=81.94) friction will be generated to stop the block from moving. Since a maximum friction that can be developed (94.75 N) is greater than the demand of 81.94 N we have more friction than needed to start the sliding action of the block.- Is my understanding correct?
@markmattsonpe I'm still confused as to when to use the correct formulas for the Moment of Inertia problems such as 1:12:16 and 1:40:00 and your other video in 2022 as well
Hi Mark. For the final question in this problem set, @ 1:30:00 the 2nd problem you have on static friction, why does the friction force (Ff) move UP the plane?? That is the only part of this problem i'm having a hard time wrapping my head around. Everything else makes sense, but in every other static friction problem with a block on an inclined plane like this, it usually is in the direction down the plane. Could you please elaborate? Thanks.
On the location of the centroid that went so awry you just added the y value for the circle incorrectly it was supposed to be 4 and not 3.5 and with that the final answer was 2.868 which seems logical and appears as an answer at 2.9
You are absolutely correct that the y value for the circle should be 4. However, the math on the circle area does not check out. As Jay in the comments pointed out, the circle area is really 28.27 mm^2. It's a great thing having friends that check the work out... it would be even better if you could bring a friend with you to the FE!
The resultant of the triangle should be 3m as written. It is 2m to get to the end of the rectangle, plus 3m/3 to get to the center of the triangular force. See 22:25.
Why do we NOT use the parallel axis theorem in the first MOI problem (I shape), but we DO use parallel axis theorem in the second MOI problem (rectangle minus two circles)?
When shapes share a common horizontal axis of symmetry (like the big box and the two voids for the I shape - they are all symmetric about their respective center points and these points all fall on the same horizontal line), you can simply add/subtract the moments of inertia. When shapes do not share a common horizontal axis of symmetry (the center of the circles are not on the same horizontal line as the center of the box), you need to use the parallel axis theorem and add the Ad^2 term as the "d" value is no longer zero. I hope that helps!
Hello! For the last question on the inclined plane, if Fx-P > friction force, then the block moves. How do you know if it will move up or down the plane? My guess is if Fx > P, then it moves downwards, but if P > Fx, then it moves up the plane. Is this correct? Thank you!
P has to be greater than Fx + ff in order to go up the plane. However, Fx has to be greater than P + ff, to go down the plane. The force of friction always resists motion. Here's a similar problem th-cam.com/video/o_1YqyJfrAA/w-d-xo.html.
For part F (The I beam question) could you explain how to use the parallel axis theorem. My brain would never think of subtracting out the voids. Every time I see moment of inertia I go straight to parallel axis thm. Great video btw!
The parallel axis theorem requires that you add all moments of inertia of the various "parts" about their centroids (that's the sum of I part of the equation) and then add the impact of the distance that "part" is from the neutral axis (that's the sum of Ad^2 part of the equation). For the I-beam, the big rectangle and the voids all share a common centroid, (they have a common horizontal axis of symmetry) so the distance in the Ad^2 part is zero. For the next problem, the rectangle with the circles, the centroid of the circles are not at the center of the big rectangle, so they deduct from the total moment of inertia. Does that help?
@@MarkMattsonPE Yes it does! Also, is this the only topic you reviewed from last year? Im looking for videos on different topics and I dont see any on your channel.
Obed! What's up! It's great to hear from you... I hope all is well. I'll be posting more FE review sessions so stay tuned and send me an email to let me know what you want to see covered.
Why is it that when you are adding moments to solve a force you do not included Ma for example (like in min 55.33) in the equation but when you are solving for that moment you do include Ma in the equation ?
Thanks for the question. My quick answer is you can't push a rope... or a cable! So in this case, the cable at the support only has horizontal and vertical forces, there is no moment at point A (Ma) for the cable problem at 55:33. However, another problem (see around 21:50) with a cantilever from a fixed or rigid support does have a moment at point A since the support and beam are rigid. Does that help?
@@MarkMattsonPE I need core subject Structural design n steel too, Geotech Water and environmental Transportation and mechanics of materials, Thanks Mark
you're so right... I put this one together way too last minute and it needed much more double checking... I'll be checking future sessions much more closely. thanks for the input.
@@MarkMattsonPE oop sorry for sounding like an asshole. Nevertheless I used your problems in my practice packet. I solved them on my own and they helped a lot. Thanks.
I do not care about the mistakes, but what I care the most is about your time and the great explanation. I really appreciate it, muchas gracias 🙏🙏🙏.
I just started taking a mechanics of materials CE class at the polytechnic school in California.. I am using this video for statics review to help prepare me. Thanks!
Hi Mark, great content. About the last question. I did force balance in x direction. My argue is that if the sum is ZERO then the body will stay at rest. If
Fx +Ff > P then body will move down. Otherwise if Fx + Ff < P then body will move up. The trick here is on what direction friction force is acting (up or down). Appreciate feedbacks.
The moment in question 2 should be 866 kN-m and not 946 kN-m. I plugged in the numbers half a dozen times exactly how you have them written and get the same answer. The final distance for a force of 226.6 kN should then be 3.82 m.
Nevermind, I paused the video 5 seconds before the mistake was caught.
Hey, thanks for the help! You are absolutely correct.
Thank you for this. The TH-cam algorithm did good sending me here.
A few mistakes, that you were clearly aware of haha, but overall very helpful. Thank you!
REALLY HELPFUL MARK MATTSON. Can't thank you enough for this!!!
Thank You! Please do more of these. Very helpful.
I have more sessions in the works and just need to get the time to go live with them. Hopefully, I'll see you at some of them!
@@MarkMattsonPE Yes, please do more sessions this one is great! I've not passed the FE 3x so 4th times the charm right?
@Mark Mattson : Regarding the last problem: " Realizing that full friction of 94.75 N does not get generated /mobilized until it is demanded by net forces is the key" In this problem only 81.94 N or slightly more (101.56-19.62=81.94) friction will be generated to stop the block from moving. Since a maximum friction that can be developed (94.75 N) is greater than the demand of 81.94 N we have more friction than needed to start the sliding action of the block.- Is my understanding correct?
Really helpful!! Thank you so much for this!
@markmattsonpe I'm still confused as to when to use the correct formulas for the Moment of Inertia problems such as 1:12:16 and 1:40:00 and your other video in 2022 as well
thank you so much ,I just saw your video and I subscribed directly ,thank you ,please do more , I want to take the FE test and that's was so helpful.
We're starting a full semester long review tomorrow night! Thanks for subscribing... I hope the additional sessions help :)
@@MarkMattsonPE thank you alot
Very helpful Mark, Thank you!
For the centroid problem E I came up with an answer of 2.56mm.
Hi Mark. For the final question in this problem set, @ 1:30:00 the 2nd problem you have on static friction, why does the friction force (Ff) move UP the plane?? That is the only part of this problem i'm having a hard time wrapping my head around. Everything else makes sense, but in every other static friction problem with a block on an inclined plane like this, it usually is in the direction down the plane. Could you please elaborate? Thanks.
Thank you so much for this :)
On the location of the centroid that went so awry you just added the y value for the circle incorrectly it was supposed to be 4 and not 3.5 and with that the final answer was 2.868 which seems logical and appears as an answer at 2.9
You are absolutely correct that the y value for the circle should be 4. However, the math on the circle area does not check out. As Jay in the comments pointed out, the circle area is really 28.27 mm^2. It's a great thing having friends that check the work out... it would be even better if you could bring a friend with you to the FE!
On question #2. when solving for the moment for the triangle the distance should be 5m not 3m. So Ma is 986kN.
The resultant of the triangle should be 3m as written. It is 2m to get to the end of the rectangle, plus 3m/3 to get to the center of the triangular force. See 22:25.
Why do we NOT use the parallel axis theorem in the first MOI problem (I shape), but we DO use parallel axis theorem in the second MOI problem (rectangle minus two circles)?
When shapes share a common horizontal axis of symmetry (like the big box and the two voids for the I shape - they are all symmetric about their respective center points and these points all fall on the same horizontal line), you can simply add/subtract the moments of inertia. When shapes do not share a common horizontal axis of symmetry (the center of the circles are not on the same horizontal line as the center of the box), you need to use the parallel axis theorem and add the Ad^2 term as the "d" value is no longer zero. I hope that helps!
The 2nd force components in red should be F2x and F2y
Thank you so much, it was very helpful
Thanks for the feedback! A new version and new problems are coming this week on 2/1/22! th-cam.com/video/VDflcZZkB30/w-d-xo.html
Hello! For the last question on the inclined plane, if Fx-P > friction force, then the block moves. How do you know if it will move up or down the plane? My guess is if Fx > P, then it moves downwards, but if P > Fx, then it moves up the plane. Is this correct? Thank you!
P has to be greater than Fx + ff in order to go up the plane. However, Fx has to be greater than P + ff, to go down the plane. The force of friction always resists motion. Here's a similar problem th-cam.com/video/o_1YqyJfrAA/w-d-xo.html.
For part F (The I beam question) could you explain how to use the parallel axis theorem. My brain would never think of subtracting out the voids. Every time I see moment of inertia I go straight to parallel axis thm. Great video btw!
The parallel axis theorem requires that you add all moments of inertia of the various "parts" about their centroids (that's the sum of I part of the equation) and then add the impact of the distance that "part" is from the neutral axis (that's the sum of Ad^2 part of the equation).
For the I-beam, the big rectangle and the voids all share a common centroid, (they have a common horizontal axis of symmetry) so the distance in the Ad^2 part is zero.
For the next problem, the rectangle with the circles, the centroid of the circles are not at the center of the big rectangle, so they deduct from the total moment of inertia. Does that help?
@@MarkMattsonPE Yes it does! Also, is this the only topic you reviewed from last year? Im looking for videos on different topics and I dont see any on your channel.
Hey professor! 👋🏾
Obed! What's up! It's great to hear from you... I hope all is well. I'll be posting more FE review sessions so stay tuned and send me an email to let me know what you want to see covered.
Must Be Higher Than Diff TimeS.. && Faster Than Diff IntellS.................Not Vice Versa.......??
Why is it that when you are adding moments to solve a force you do not included Ma for example (like in min 55.33) in the equation but when you are solving for that moment you do include Ma in the equation ?
Thanks for the question. My quick answer is you can't push a rope... or a cable! So in this case, the cable at the support only has horizontal and vertical forces, there is no moment at point A (Ma) for the cable problem at 55:33. However, another problem (see around 21:50) with a cantilever from a fixed or rigid support does have a moment at point A since the support and beam are rigid. Does that help?
Please make more videos as you can do ,
I'm planning out more videos... please let me know what you want covered. Thanks!
@@MarkMattsonPE I need core subject Structural design n steel too,
Geotech
Water and environmental
Transportation and mechanics of materials,
Thanks Mark
@@MarkMattsonPE Can you please do a review on dynamics? Thank you
bro so many mistakes
you're so right... I put this one together way too last minute and it needed much more double checking... I'll be checking future sessions much more closely. thanks for the input.
@@MarkMattsonPE oop sorry for sounding like an asshole. Nevertheless I used your problems in my practice packet. I solved them on my own and they helped a lot. Thanks.