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(II) For two blocks, connected by a cord and sliding downthe incline shown in Fig. 34 (see Problem 20$)$ , describe themotion $(a)$ if $\mu_{A}<\mu_{B},$ and $(b)$ if $\mu_{A}>\mu_{B} .(c)$ Determinea formula for the acceleration of each block and the tension$F_{\mathrm{T}}$ in the cord in terms of $m_{\mathrm{A}}, m_{\mathrm{B}},$ and $\theta ;$ interpret yourresults in light of your answers to $(a)$ and $(b) .$

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a. A will pull B along.b. See image for solutionc. See image for solution

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

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

McMaster University

Lectures

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

11:48

(II) Two blocks made of di…

16:54

19:28

Two blocks made of differe…

06:35

05:37

'Two blocks made of d…

01:08

A cord is wrapped around a…

07:54

03:46

Problem: Two blocks, made …

06:27

06:41

05:30

Consider two blocks connec…

02:54

Two blocks of masses m and…

15:14

A block of mass $m_{1}$ sl…

for this problem. On the topic of friction, circular motion and drag forces, we're told that two blocks connected by a cord are sliding down. The inclined plane has shown in diagram you want to describe the motion. Firstly, if new A is less than you be, revenue is the coefficient of friction. And then secondly, if New A is greater than newbie, and then to determine a formula for the acceleration of each block and the tension in record in terms of the massive a. M a, the massive B M B and the angle theta, we don't want to interpret the results in light of the answers to A and B. So, firstly, for question A. If mu A is less than you be, the untethered acceleration of M A would be greater than that of M B. If there were no called connecting the masses and they would run away from B and B if they are joined together, May would be restrained by the tension in the cord. An M B would be pulled forward by the tension in that card, and the two masses would have the same acceleration now for Part B. If new A is greater than maybe than the untethered acceleration of M A would be less than that of M. B. So even if there is a card between them, MB will move closer to M A. And there will be no tension in the cord if the inclined plane or long enough, eventually, MB would catch up to Emma and begin to push it down the plane. Now for part C, we want to know. We want to examine a formula for the acceleration of each block and the tension in the quad. So if we assume that new A is less, then newbie, then we have analysis some similar to problem 20. So if we look at problem 20 we have the acceleration. A can be written as g into the mess em, eh into sign data minus new a cosign theta plus and be into scientists to minus um, you be course I indeed, to and all of this divided by the sum of the masses m A plus MB. No, Obviously, from Newton's second law, we can see that m a times A is equal to m a times G scientific to minus new a m. A G call sign of theater minus the tension FT. Which means that the tension force FT. Is equal to M a times G time scientific data minus new A m A G call sign of theta minus m eight and acceleration A and we can write this as m A G science data minus, um, you a m a g co sign of theater minus m A. And we can replace the acceleration with the expression for the acceleration above, which is G into m A into sign data minus new a co sign data plus and be into sign data minus newbie close enough to to all of this divided by and some of the messes m a plus MB. So if we simplify this expression, we get this to be m a times m b times G call sign of data divided by m a smb into the difference in the coefficients of friction New B minus new A So we found the acceleration and now we have the tension force f t. So if we do, if you repeat this for now, um, you a greater then you be and we can follow the analysis again of the problem Number 20 but not include the tension forces. So each block has its own acceleration. And for block A, we know that the sum of all the forces in the vertical direction on Block A is firstly the normal force on block A minus m a g call sign of theater and this was equal to zero, which means that the normal force acting on Block A f A is equal to M A g co sign of theater. So if we do the same for the forces along the X direction in some of these, if X A is equal to m a g sign of data minus the frictional force acting on block A f if our unblock a must equal to the massive block wait times acceleration M A by Newton's second law. So we can see that m a times A is equal to m a G scientists to minus the frictional force which is the coefficient of kinetic friction. You a times the normal force f and eight And we can actually write this as m a times G sign of data minus mu A and we can write the normal force as from the expression above M A G co sign of theater, which that means we can find the acceleration of Block A to B G into sign data minus mu a co sign of theater. So we have the acceleration for block alone. Now the acceleration for Block B is as follows again. We take the some of the forces acting on Block y along the Y direction F Y B, and this is equal to the normal force. Acting on B f N B minus m b g call sign of data, and this is equal to zero. So this means that the normal force acting on Block B F N B is equal to M b times g times, the core side of theater. And again, we need to do the same for the forces acting on Block B along the extra action. So if some of the forces in the X direction Block B is m A, actually be so m b, she scientists to minus the frictional force acting. That's the frictional force acting on Block B. And this was equal to the massive B times its acceleration M b times A B and therefore we get M b times a B to be equal to M B times G scientific data minus the frictional force on B which we can write the coefficient of kinetic friction newbie times, the normal force, F N B. And we can write this as M B G signed Peter minus um, you be and we can write the frictional force and be as M B G call sign of theater, which means we can get the acceleration of block B A B to be G into science data minus mu. Be co sign data. Note that since me away is greater than you be like we said above and we have acceleration of a is greater than the acceleration of me. And we have the tension force in this scenario. FT to be zero

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