Colin Selleck

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@valeriajackson9395 Күн бұрын

Why is the displacement in yb = -64?

@ColinSelleck Күн бұрын

The origin is at the jump point. So y is negative. it is -4 - (3/5)*100 = -64 m.

@valeriajackson9395 Күн бұрын

@@ColinSelleck Dear Mr. Selleck, just know that you have a place in heaven, infinites thank you 🙏

@valeriajackson9395 Күн бұрын

In minute 7:38, why is the x-component of Va = Vacos25 instead of Vasin25? Given the orientation of the right triangle.

@ColinSelleck Күн бұрын

You are correct. I solved the problem assuming the angle was measured from the x axis, which is what I believe was intended.

@valeriajackson9395 Күн бұрын

@@ColinSelleck Makes a lot of sense, thank you so much!

@valeriajackson9395 9 күн бұрын

Hello Sir! Thank you so much for this lectures. I went from zero to hero with your explanations in dynamics.

@ColinSelleck 9 күн бұрын

You're welcome!

@atiyamolla6330 10 күн бұрын

For 15:30 I don't understand how we found uA

@ColinSelleck 9 күн бұрын

The vector uA is parallel to the y axis so it is 0i+1j+0k.

@tracyradsvick3450 Ай бұрын

This would be more helpful if there was any verbal/audio/written instruction and not just you showing it. Maybe edit the video and add audio to it?

@HOSSAINSHWAJTAJRIANSTUDENT 3 ай бұрын

5:05

@philipterzian4581 6 ай бұрын

FYI that's David Acheson (1921-2018), son of Dean Acheson, on Feynman's right. He was a prominent Washington lawyer, ex-government official, and of course, member of the Rogers Commission.

@yigitcan824 10 ай бұрын

Professor I have a question here what' the difference between centroid and center of gravity?

@ColinSelleck 10 ай бұрын

They are the same for a homogeneous body (constant density). Other than that, they are different.

@yigitcan824 10 ай бұрын

Professor, I have a question here.What tilda means actually like at 1:38 ?

@yigitcan824 10 ай бұрын

:Professor I wonder sth at 18:09 why maximum bending moment occurs when the shear force is zero?

@ColinSelleck 10 ай бұрын

Do you have the Hibbeler textbook? Look at chapter 7.3 where it defines dM/dx = V. So when shear is zero the slope of the moment diagram is 0, so you are at a local maximum in the M -x plot.

@ColinSelleck 10 ай бұрын

Also see engineeringstatics.org/VM_relations.html#:~:text=The%20slope%20of%20the%20moment%20diagram%20at%20a%20point%20is,moment%20diagram%20and%20vice%2Dversa.&text=Change%20in%20the%20moment%20between,area%20under%20a%20shear%20curve.

@yigitcan824 10 ай бұрын

@@ColinSelleck Oh thanks a lot professor :)one more thing, forces in the pin,for example, are internal forces right?We do not show them in the entire body diagram

@ColinSelleck 10 ай бұрын

@@yigitcan824 The pin forces are not internal forces. They are support reactions and need to be solved for before you start on the internal forces.

@HashemAljifri515 10 ай бұрын

In 12.40: took moment about wheel B and got NA = 470, NB,570. Any suggestions? If we were to take the moment about any point we should get the same answer ..from statics course

@ColinSelleck 10 ай бұрын

I made a mistake in the video Na is 470 and Nb is 571.

@yigitcan824 10 ай бұрын

Professor I do not understand the logic of sign convention at 5:11 , could you explain me a little bit more ?

@ColinSelleck 10 ай бұрын

When you draw a FBD of the left side of a beam, put the positive normal force to the right, the shear force down, and the moment CCW. Invert the directions when you draw a FBD of the right side of a beam.

@yigitcan824 10 ай бұрын

Professor I have a question, at 4:02 there're 2 trusses with tensile and compressive force and I got that.But I wonder what aplly these tension or compressive forces?I mean For example in the first truss, member is under tensile force from both ends but from where these forces coming?I'd really apreciate it if you could explain me.

@ColinSelleck 10 ай бұрын

The forces come exclusively from the loads on the truss (snow, cars, peoople, ...). We ignore the wight of the truss members because it is usually much smaller than the loads. So in an unloaded truss, the members experience no forces.

@yigitcan824 10 ай бұрын

Professor I have a question.In equilibrium in 3D, sum of the moments about a point or an axis must need to be zero? Or maybe both of them?I'd be grateful if you could help me to understand it

@ColinSelleck 10 ай бұрын

Yes, in equilibrium, the sum of moments is zero. You're always summing moments about an axis, even in 2D you're summing moments about the z axis that passes through a poin in the xy plane.

@daddyecon 10 ай бұрын

Hi, very informative video! Do you know if there is any way to add an attempt limit to the password so students aren’t inclined to try and guess it? Thank you!

@aedemirel_ 11 ай бұрын

15:35 Why Ua is the positive y direction. How can we understand that ? Maybe it is negative. Then it would be still on the a-a axis. I cant understand that.

@ColinSelleck 11 ай бұрын

The unit vector can be in either direction. The answers will just be the negative of each other. The magnitude will be the same.

@clearflow7925 11 ай бұрын

8:06 why is the IC there?!?!?!?!

@ColinSelleck 11 ай бұрын

Because the gear rolls without slip and the contact point on the rack B is not moving.

@haraldisdead Жыл бұрын

HILF brought me here.

@michaelo9235 Жыл бұрын

just hope you know, you're our hear saving my life right now. I wish my dynamics proff was as good as you

@ColinSelleck Жыл бұрын

Glad to help!

@wheatlysparble7900 Жыл бұрын

Thankyou so much. Your tutorials are very helpful

@ColinSelleck Жыл бұрын

You're welcome!

@ajwhite5471 Жыл бұрын

so, the last example would s ( in 2-s-s dot) be constant, as in zero??? or what would s be ??/

@ajwhite5471 Жыл бұрын

what was theta initial?? how did you find theta???

@ColinSelleck Жыл бұрын

TIme stamp?

@ajwhite5471 Жыл бұрын

@@ColinSelleck theta o, is it zero

@ColinSelleck Жыл бұрын

@@ajwhite5471 At 19:30, theta 0 is 0.

@LucasR142 Жыл бұрын

hello thanks for the video, why does gravity not do work in the second example along with the 5ft/s force?

@ColinSelleck Жыл бұрын

I believe I did include gravity. And the 5 ft/s is not a force, it's a speed.

@kevinbruns2771 Жыл бұрын

Thanks a lot!

@ColinSelleck Жыл бұрын

You're welcome!

@MegaDanielChannel Жыл бұрын

why is Y tilda 3.0 for C in example1 and not 2.0?

@MegaDanielChannel Жыл бұрын

like why do we go an extra 1/3 on the y axis for segment C but didn't do the same for segment A, where we went half the y distance

@ColinSelleck Жыл бұрын

y tilde for segment C is 0.5 + 2 + 1.5/3 = 3 m.

@taayanewmanshivolo4282 Жыл бұрын

Hello sir, isnt the distance that the weight of the box is doing supposed to be from the bottom to the centre point G? (4sin(60)+1)

@ColinSelleck Жыл бұрын

No. It's 4 sin(60). The center of gravity ends up 1 foot above the floor.

@reprogrammedtohae4286 2 жыл бұрын

at 5:48, where'd you get the 1/sqrt13? Lol

@ColinSelleck 2 жыл бұрын

Oops. That would be sqrt(10).

@ItsRanked 2 жыл бұрын

5:28 you said any force along the y axis but Mo is not along the y axis

@ColinSelleck 2 жыл бұрын

But there is a component of Mo along the y axis.

@ItsRanked 2 жыл бұрын

@@ColinSelleck so what you want is that you take the moment of any point that has a component over the y axis then do the dot product to get this component?

@ColinSelleck 2 жыл бұрын

@@ItsRanked Yes. It's called triple scalar product.

@meta196 2 жыл бұрын

All forces are zero in determinate structures, how does this FBD help?

@ColinSelleck 2 жыл бұрын

Time stamp?

@meta196 2 жыл бұрын

3:24 is there a mistake here? Why is Y axis taken at an inclination?

@ColinSelleck 2 жыл бұрын

I don't see a y axis at 3:24.

@ajwhite5471 2 жыл бұрын

why didnt you do the kinetic diagram for the second example and explain the why they are in those direction

@ItsRanked 2 жыл бұрын

I did not understand what is physically the perpendicular component here and why Fperp + F par = F

@ColinSelleck 2 жыл бұрын

It's from the Pythagorean theorem and it's F = sqrt(Fperp**2 + Fpar**2)

@mehdihani2932 2 жыл бұрын

why in semi circle X tilda zeRo?

@ColinSelleck 2 жыл бұрын

Because the semi-circle centroid is on the y-axis.

@mehdihani2932 2 жыл бұрын

sir pls upload chapter 10 too

@ItsRanked 2 жыл бұрын

Thank you so much

@ColinSelleck 2 жыл бұрын

You're welcome!

@matovumatthew97 2 жыл бұрын

Try to be louder next tym plizz

@younamsayin 2 жыл бұрын

At 14:40, could you please explain why the acceleration a(G) is towards the right? Shouldn't it be towards the left since the angular acceleration is CCW?

@ColinSelleck 2 жыл бұрын

The pipe has a angular acceleration wrt to the truck that is CCW. But the mass center of the pipe has an acceleration to the right.

@younamsayin 2 жыл бұрын

@@ColinSelleck But shouldn't the acceleration of the mass centre be consistent with the angular acceleration?

@ColinSelleck 2 жыл бұрын

@@younamsayin No. The alpha is wrt to truck and it can be CCW but the mass center can be accelerating to the right. Look at the relative acceleration equation.

@younamsayin 2 жыл бұрын

@@ColinSelleck i'm pretty confused here... I thought the motion of this truck will cause the ring to move towards the left. IF we were to use angular velocity instead of angular acceleration, and velocity instead of acceleration, then in this scenario, would the angular velocity become CCW and the velocity to the left?

@ColinSelleck 2 жыл бұрын

@@younamsayin The pipe does move to the left wrt to the truck. But the absolute acceleration of the ring is to the right because the truck is accelerating.

@madisonseti1335 2 жыл бұрын

In the last example, the moment caused by the force would be clockwise. Why is it positive instead of negative?

@ColinSelleck 2 жыл бұрын

I should have made it negative, then omega 2 would be negative or CW.

@madisonseti1335 2 жыл бұрын

The final example is calculated incorrectly in the very last step. Alpha is incorrect, leading to an incorrect acceleration. Minor math error.

@ColinSelleck 2 жыл бұрын

You are correct. alpha is 40.8 r/s*s and acceleration of A is correct at -12.5 m/s*s.

@madisonseti1335 2 жыл бұрын

@@ColinSelleck Thank you for these videos! I'm reviewing for my dynamics exam today and these have been an amazing refresher!

@ColinSelleck 2 жыл бұрын

@@madisonseti1335 You're welcome! Good luck on your exam.

@mehdihani2932 2 жыл бұрын

dear sir pls upload mechanics of materials by rc hibbeler lectures

@ColinSelleck 2 жыл бұрын

Sorry, I've never taught that class.

@Doomzone Жыл бұрын

@@ColinSelleck 🤯

@oskarimioek8544 2 жыл бұрын

Thanks professor! Your work is a great help

@ColinSelleck 2 жыл бұрын

You're welcome!

@ColinSelleck 2 жыл бұрын

sqrt(249.7/49.69)

@sandman69696 2 жыл бұрын

may I ask how did you get the 2.24 rad/s?

@anasabdelfattah5285 2 жыл бұрын

In 14:00 while calculating the moment about point B, why you took the moment of inertia of G mass center not C ?

@ColinSelleck 2 жыл бұрын

When you sum moments about a point other than the mass center (in this case b) you need this equation: sum moments about b = Moment of inertia about g (mass center) * alpha + vector between g and b cross mass * acceleration of mass center.