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You are trying to raise a bicycle wheel of mass $m$ and radius $R$ up over a curb of height $h .$ To do this, you apply a horizontal force $\vec{F}$ (Fig. $10.81 ) .$ What is the smallest magnitude of the force $\vec{F}$ that will succeed in raising the wheel onto the curb when the force is applied (a) at the center of the wheel, and (b) at the top of the wheel? (c) In which case is less force required?

a) $F=m g\left(\frac{\sqrt{2 R h-h^{2}}}{R-h}\right)$b) $F=m g\left(\frac{\sqrt{2 R h-h^{2}}}{2 R-h}\right)$c) From the above two results it is clear that the force required at the top of the wheel is less than the force applied at the center of the wheel.

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

Hope College

University of Sheffield

McMaster University

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Okay, so we know the tour is equal to perpendicular this and times their force. So their whole have talents in hotel to know. Have you force after, Hence, the cover *** this and of the four force After you go to the graphic intensity primitive addition off the gravity. Which city here. So therefore, we know that the program, because Edison off the gravity de can make with the square root are square minus our square. I'm sorry. No. Ah. Minus eight square. Okay. And this will give us square. Oh, are square finest are square minus two Rh class age square, which will have square root to our age minus age square. So level have four f ar minus h. You see OMG square root to our age minus age square. Okay, so therefore the four steps apply at the center of the will, uh, is equal to MGI Times Square. Oh, to r h. Mine is a square over ar minus h. And full question be well, now we know you perpendicular distance off the force. There was a pie at a top of the wheel. She'll be You got to, uh, two R minus h o K C is now even consider the diameter of the well wishes to R and I minus the the Heights. So therefore have social going on here, which is the force that was apply at the top of the wheel times the perpendicular distance off such force, which is true ar minus H, is in virtually gravity. Test the purple and do for the distance off the gravity. Which will we got it from the last question which is square root, um, to our age minus eight square. So there will have the force that was apply at the top of the wheel is just incredible to m g Times Square room two are h minus H square over to our minus H and questions he was asking us which forces less well. So now we have the force at a different position. So, as you can tell, and the values of the variables of Catholic denominator for both forces saying which is mg Times Square to R. H minus a square. But enough. The variables below the denominators difference just, uh, in case a they variable below. The denominators are my age. But in case B, the variable below the innovators to our minds stage. So therefore, we know that two ar minus age is greater than our minds age. Things to ours is greater than our Which means that the force it was a pie at the top of center. This is F B. It's Lastly, the four steps apply at a center. Okay. Is it because, um, whatever has the last number below the denominator, where is greater? So as organs help the force A as, ah, lower value below the denominator, which means foresees greater. Okay, let me let me right down to show you guys. Okay, so I that be easy with energy. It has square rule to R h. When is a square over to our by this age? Smaller than m g square root twice. Mine is a square over arm on his age. Okay, let's take a look here. So, uh, therefore, the force that was applying a tablet wheel is less so. They said at the top of the wheel, and this is the answer for these questions.

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