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(II) $(a)$ Suppose the coefficient of kinetic friction between$m_{\mathrm{A}}$ and the plane in Fig. 38 is $\mu_{\mathrm{k}}=0.15,$ and that$m_{\mathrm{A}}=m_{\mathrm{B}}=2.7 \mathrm{kg} .$ As $m_{\mathrm{B}}$ moves down, determine themagnitude of the acceleration of $m_{\mathrm{A}}$ and $m_{\mathrm{B}},$ given$\theta=34^{\circ} .(b)$ What smallest value of $\mu_{\mathrm{k}}$ will keep the systemfrom accelerating?

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a. $a=1.55 \mathrm{m} / \mathrm{s}^{2}$b. $\mu=0.53$

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Quy?T N.

October 1, 2021

(a) Suppose the coefficient of kinetic friction between mA and the plane is ???????? = 0.15, and that mA = mB = 2.7 kg. As mB moves down, determine the magnitude of the acceleration of mA and mB , given ?= 34°. (b) What smallest value of ? will keep the s

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University of Michigan - Ann Arbor

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University of Sheffield

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So in this problem, the box, Uh, it is on the plane is going up or which is on the slope is going up. And then, uh, box m B is going down so we can draw the free body diagram off these two. So I've already drawn that beforehand. Um, this slope is making angle. Dada with the horizontal now for the components. Um, we can't choose our exes according to our convenience. So exes. Ah, parallel to the plane. All the slope. And why is for Ben Dickler to the slope? So, um, the component off m a g on why a will be eso business data. So this quantity will be mm. Eight times g co sign data and the component on X access. Let's actually use Red Arrow for that. So yeah, these are on related to M A. G. So this is what this one will be a g sci fi. Also, there will be the normal force, which is acting vertically, upwards. Um, along. Why access then? Since the boxes going up, there will be frictional force which is going down, and this is we call it f off a far. Then the tension in the string will be f time Steve, which is going off similarly for ah, the box, which is going vertically down, will have mbd, which is acting downwards, and f t will be acting upwards. And in this case, we're taking ah vertically downward direction as positive. Why access? So we know right down the questions associated with all those, um, so that's using Newton's second law. Let's first tried the force on this part where the box is vertically going downwards. Um, so f off Why b is equal to and B g minus. If off t now, this whole system is going down with an ax. Elation A So that's M B times A. And from here we can see that if off dear the tension The string is M B times G minus and the times eight then, um, for the slow we have if off x a now for X A. We have detention force. We have the component of mg and we have f r. So combining all of them together we have f d minus. So, since we considered upward direction as positive so we have posited, then we have negative f loud and negatives And me G signs Vega that it's gonna be and a G signed data minus 1/4. If our which is equal to in a times now Ah, for wide election, we have f off. Why a It is equal to, um, the normal force and someone in the mg again. So we'll have if And since we considered ah, the upward direction is positive. So f and minus m a g co signed data and since there is no movement in their direction, well said, that equals zero. And from here we can see that f n is equal. Do m a g cool science data Also, um, we have to remember that since it's ah kind of dick friction, we can set f off fr equal to mu k times f off in. So yeah, that's that's one relation that we need toe, remember? So what we're going to do is we'll take the information from the two white questions and substitute that into X equation to solve for the acceleration will take Ah, these three questions, this one and this one better sort associated with y. And then we'll substitute these two questions on X um so to do so. And it s so if we do that, we see that we'll have m b times G minus m b times a minus M eight times g Scient, ada minus nooky times m a G course in theater equal to m a things. And from here, we'll see that a Zeke will do m b gene minus m a g signs it up minus m a g nu key g co sign data. Um, see that Me? Check it for doing everything correctly divided by m A plus and beets. So we have MBG We have images, scientist A Than we have mu k G uh, m a. Okay, so we don't need this g right. And ah, we use the given values. We see that, um, we have, first of all, half off G one minus science data minus mu k co sign data so they die is given in our case and mucus also given. So that's gonna be half off. 9.8 meter per second squared, one minus sign. 34 degrees, minus 0.1 fight course sign. 34 degrees, which is 1.6 meters per second squared s so what we did here is first we substituted if and in this equation toe get fr. And once we got a far, we blood that over here, then we combined. Ah. So then we have s o. We can substitute ft from this a question and this equation. And then we combine these two questions to get ah, the acceleration. Now in part B, we need to find out the smallest value of UK to keep the system from accelerating. To do so, we'll have Ah, we see that the taxation should be zero. So that means that the expression that we derived earlier, which is ah e equals half of G one minus sign Seda minus mu Keiko Science data. This whole expression should be zero so that there's an expiration. And if we do so, we'll have one minus one minus sign data minus mu K co sign Data is equal zero. From there, we see that new K is one minus. Sign data over co scientist. Now we know they da is, um, 34 degrees. So that's one minus sign. 34 degrees divided by co sign 34 combining everything together. We see that UK is 0.53 Thank you.

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