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FIU minLOMRQuestion 11ptsConsider a "card bridge" shown in the image below: This is structure built with playing cards; like house of cards, but simpler;(...

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FIU minLOMRQuestion 11ptsConsider a "card bridge" shown in the image below: This is structure built with playing cards; like house of cards, but simpler;(Each of the straight; black line segments in the image represents _ playing card ) Each of the five cards has the same mass; Also; each of the four cards in the base Is inclined atan angle 0 with respect to the horizontal; as shown in the figure The coefficient of static friction between the cards and the table Is 0.352 There is effec

FIU minLOMR Question 11 pts Consider a "card bridge" shown in the image below: This is structure built with playing cards; like house of cards, but simpler; (Each of the straight; black line segments in the image represents _ playing card ) Each of the five cards has the same mass; Also; each of the four cards in the base Is inclined atan angle 0 with respect to the horizontal; as shown in the figure The coefficient of static friction between the cards and the table Is 0.352 There is effectively no friction between the cards and other cards. What is the minimum 0 can be for this card bridge to be stable? (Give your answer in degrees )



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Prove the following identities. Assume that $\mathbf{u}, \mathbf{v}, \mathbf{w}$ and $\mathbf{x}$ are nonzero vectors in $\mathbb{R}^{3}$. $$(\mathbf{u} \times \mathbf{v}) \cdot(\mathbf{w} \times \mathbf{x})=(\mathbf{u} \cdot \mathbf{w})(\mathbf{v} \cdot \mathbf{x})-(\mathbf{u} \cdot \mathbf{x})(\mathbf{v} \cdot \mathbf{w})$$

Support. So for this problem, behalf mg have signed often out. It's equal to new times and and you for N is equal to, um G ham school sign If I know So we have mg. I was assigned. I find out it's going to new accounts. Am aji Cool sign or not, we can cancel out having Jay here. So hand int I know is they go to you so I not it's going to universe hundreds of meal.

In the given problem, there is an inclined plane because having no friction in the beginning, if he's but in blank man was amble off. Inclination is Tetum. There is must less and frictionless pretty. But what it from there our stream exposing There is a mosque, em a pledge because being and there are two blocks over that incline having masses m and um these are also attached with the help off shrink. No, If you look for all the forces acting over these blocks, this is the weed off the and you look capital N g for Rich. This is the tension Hey do in the string. And here this will also get attention. Tito No, no paid off this block. Yes, you and G acting vertically don't work. Who's component perpendicular to the plane will be two mg Costinha and along the plane downward direction It will be too. MGI sign Peter Mary's attention. The one in the string connecting the blocks M and to him No, we look for the G eight off the block off. Mask him, which is N g. Acting vertically don't work. So just component will be M g course teeter and here. This will be MGI sign Pita. So this is the force diagram off the system given here from the question? No. Well, man, the free body that rounds off different blocks, we will consider these people Bagram's you park. First off the problem taking F B D free body diagram off block, hanging block off muss. I m Yet MGI is equal toe Tito because the system is in equilibrium. So there should be no net force acting on the blocks. That's why this mgs quantity toe now cooking fbt off block off. Master him F b d off block off Maas do in we get the tension. People in the trip is equal to tow n g sign pick up again because there is no net motion in this block, So there should be no net force acting on it. Gonna be taking the free body diagram off block off Maas. I m we get de even you don't word addiction. Mrs. Stephen Bless MGI Signed Tita is equal. Do Tito again because there is no no portion in this block also. So finally we can put the value of the Stephen here So we get Don't MGI sign. Tita. Yes, MG Scientific data as digital. So you two comes out to be three MGI signed. Pick up and using this aggression The first equation M g is equal to p toe means three mg signed Peter. So the answer for first part mass off this hanging block is equal to three. I m sign Pita. This is the answer for the first part. Now, for the second part, we have to find all the stations in the strings which are even anti toe. So we already know the tension. Tito, which waas simply three mg scientist. Oh, So NATO is three MGI scientific and for even in order to find even This is the answer off even already. We're having even is two mg sign Peter. So these are the two answers for the second part off the problem? No. You just given that the mosque? Oh, the hanging law, I suppose Toby devise offered initial moss and all. We have to find acceleration off all the blows in part C. So as the Mars of the block waas three m signed Peter so on making it twice it will become too into 3 a.m. sign pickup means six m signed pita. So now taking the free body diagram as the mask has been double. So this lock will spark falling down. Suppose with an acceleration A So these blocks will start moving up the incline with the same accelerations it So taking the free body diagram off this hanging block we get this AM dish G minus. Tito is equal to em. Dash or we can see six and G Signed Peter, this becomes she extended. You sign Pick up minus They do is equal to 6 a.m. Sign pita. We can make it a question. Number one now taking the free body diagram of the block off this master when we will get Stephen minus two in just 90 de even minus two mg sign Tita is equal to two m a. So it gives us even is equal toe mg sign Kita Plus two and a love you dick Free body diagram off this block off Mass M as this one is also moving upward So tension Tito will be more than the addition off even and mg scientist. So the net force acting on the block off mass m really human ask Tito minus Stephen minus mg signed Tito which will come out to be equal toe Emmy. So from here, this Tito will become the one unless MGI scientific unless m e And here we can put the value of the Stephen. Yeah, so he to finally becomes do n g sign Pekka Plus to an A plus and g sign Peter bless any So finally began Say it becomes three mg scientific data plus three and me Now finally we put this value off Tito in equation number one putting the value of need Oh, no question Number one We get to know this Pass six MGI sign Pita, My last Ito And in place of Tito Tree mg signed Tita minus three Emmy is equal to six and sign teeter So finally just becomes three m g sign pita is equal to three Any less six and scientists you doing can't business, So the acceleration e can be taken out as a woman. So we'll get three m plus 6 a.m. signed Tita into a so acceleration will be given asked three mg sign pita divided by 3 a.m. which can be taken as a woman out remaining inside will be one less so science pita that this tree and can be cancer. So finally, acceleration off all the blocks comes out to be G signed Tita divided by one plus Do sign Tita, this is the answer for part C No, for part B, we have to find the tensions in the streets. And these tensions are for even do you want? He's given us who am design Peter, Let's do any Do MGI signed Peter less too. I am a on in place off a We put this value G scientists are divided by one tlas two scientists and then being take this OMG is apartment out? We get sign Pita Yes, sign pickup. Or you should have taken the scientist also as a common Oh, but just to find teacup. So finally this answer we come out Toby do MGI scientific data one glass one upon one plus to assign Peter Oh two mg sine theta After peaking as him month less you will sign Pita less one invited by one plus Neuroscientist Donna And finally this tension Steven comes out to be four mg sign pita one plus scientific data divided by one plus don't sign. Take up. This is the answer for the ancient even now for tension. Tito, we put the value off acceleration a here intention to do which was three m d Santa lost. Really? So it is three MGI scientific data Last three I m a. And again in close off a we will put g sign Peter divided by one plus two. Sign Tito. So taking this three mg scientist again as a formal out we get one glass one by one, plus two sign pita or three mg Signed Peter are protecting LCN one plus new scientist plus one divided by one plus two Simon Peter. This will become one plus 12 and this to also and be taken out. So p two equals 26 mg signed Tita one bless scientist up divided by one plus who will sign Peter. So these are the two answers off the fourth part of the problem. Now, in order to solve the next parts off the problem, we have to consider the coefficient of static friction Also between the blocks off mass to him and them, and between my inclined surface and disgraceful, gentle, static friction he's given asked us. So we have to find a maximum and minimum possible value off mass off, hanging long so that the system nearly name in equilibrium. So, first of all, we have to find a maximum possible value off the mask off this hanging block. In order to get this the block, this hanging blob will be having a tendency to move down. In order to get the maximum value of the mass, the tendency will be to move down. As the tendency of this block is hanging. Locker is to move down. So these blocks, they have to move up a long time climb plane. So as the mass M is moving up, so there will be a force of friction. F one acting on this law and the force of friction f to acting on this law. Between this block and up inclined surface, the mosque is Guim. The weeds already be known. OMG the home run ends two mg cost Eder as this endless Tita, this was take up the normal reaction This time we will have to consider No, I m g because Tita here, this and one equals visual. Considered this s f one This force of friction at one and there this should be after. And the mosque of this block is mg So And this is MGI course, Tita. And this normal reaction and do as mg cost Peter. No, The components along the inclined surfaces are to enjoy a sign. Peter. And, um do you sign up? This is the ancient even and even And here is component this toe. This is simply m g scientist as they have joining the first today in Mosul. I believe that the only difference is the forces or friction have also been included here. So F one will become, um, us into anyone. Or we can see us into do O M G cause Tita and F toe will become new s in the end to arm us into MGI cost Tita. So these are the two forces of friction. The shock on new inclusion in the right. So as the system is in equilibrium, So taking back William off this hanging law, this video should be equal toe mg must fall here. We are getting the value off the maximum possible mass with this hanging Look, This is fifth part of the problem. So to solve this this Tito is well OMG taking the free body diagram off this hanging block now taking the free body diagram off this mask m we get they don't He's gonna tow the one bless mg signed Tita. Unless the force of friction after which comes out to be new s in tow MGI course it does Now taking the free body diagram off this block. Having master him, we get to know this Stephen is equal to toe MGI signed Pita bless F one left one You know this is us into two mg Well, cedar No putting the value of this Stephen here in this equation for Peter, he took himself to be do MGI scientific les do into new s MGI cost Tita bless mg sign Peter bless new s MGI costed Oh, or combining them we get it becomes three mg scientific data plus three new s MGI Of course Tedo This is the value off. Hey, To so the maximum possible mass here comes out to be MGI Is it okay to And who is this? We can take this three mg common leaving behind scientist a plus U s costs eat up this GBP cancer. So the maximum possible mass off the hanging block at which the system may remain in equilibrium is three m scientific data plus new s. Costinha. So this is a far this is the answer for fifth part. Now we have to find the minimum possible force. The minimum possible marks off this hanging law that people have to change the diagram again. Because in case off the minimum possible mass off the block this hanging block the same object we will consider. It's block this hanging law to be moving upward in case off the minimum mass meeting on possible mass. It was a star moving up as this will be moving up, this will be having our tendency to move up. Not exactly moving this block. These two blocks will start moving down here. They start sliding down as they are sliding down. So the forces of friction now will act in upward direction. This left Oh, and this effort, the store, all the things that women saying whether it is, is OMG on this mg distension d do Here. This is also t to the Stephen and this Stephen this component two mg cost. Tita, this is Tita. This gunman in OMG Scientific data this component mg Costinha and this combo nen mg sign pita. All the things we've been seeing No taking again the free body diagram of this union block. You know MGI is going toe to toe. So in order to find the value of this m the minimum possible mass off the hanging globe. Most of all we need to find the Streeto is in order to find this, Tito will consider the free body diagram off the body off the mastering so it will become Stephen Bless f one is it Porto MG signed Teeter What? We can see even less new ists in the room MGI cause Tito is going toe to o m g Signed Peter So even comes out to be you'll MGI scientists are minors um us in tow to mg costing up? No, We will consider the free border Dagenham off this block off mass m So here it's a stable which is equal to d two plus after it will be and it will be equal toe anyone plus m g signed tickle So take a is equal to de even plus m g Santita minus f two. What we can say in place off even you know So mg signed Peter minus newest into true mg Costinha, then plus m g scientific data and minus for effort to it becomes new s m G cost eater. So finally this Tito comes out to be OMG Santita less energy, scientific up news three mg signed Tita and minus yes diplomatic Rosita minus messenger Casita it becomes three us mg costs. Eat up or we can see m G three and G can be taken out for Ung Scientific data minus us. Course he did so using this equation, M G is equal to three MGI scientific data minus numerous Bastida. So finally the minimum mass possible for the hanging object is three m signed Tita minus Mu s cost pita. So this is an answer for the fifth part of the problem. Now, for the final part of the problem, we have to find the difference between that oceans ancient t two in the case off Maximus and the minimum mask off the hanging law. So simply maybe subject Tito Max and even next and they will come out to be three mg will be taken. Haman outsourced. Signed Peter Plus new s, Costinha. Miners. Scientific data. It will become plus us, Costinha. So finally, the answer is six MGI in the US into cost it up. So this is the difference in the tensions in the strains he do, Max. Minus Stephen. Max. Thank you.

Once again, welcome to a new problem. This time we are dealing with forces. And when you think about Newton's Newton's second law, what tends to happen is that force is the product of the mass and the acceleration. So that's Newton's second law. And uh there's whenever there is equilibrium and we're thinking about equilibrium in terms of uh let's say you have a box that's sitting on the table and the box has a weight MG. And uh the wait is opposed by the reaction off the box. So this is Newton's The 3rd law where we say action and reaction force is action and reaction force is equal uh equal and opposite. So action and reaction forces are equal and opposite. So mutants. Third law helps us with equilibrium. Where we're saying the summer forces in the X direction is zero and the summer forces in the Y. Direction is also zero. And that tends to happen whenever there is equilibrium. So in this new problem, we have um we have uh an incline, Okay, we have an incline and it's shipped such that there's a block on both directions. So this is this is oh incline. And On the incline their seats two boxes, the first box sits uh right here. And then the second box sits on this other side like that. The first box has a mass M. One and then the other one has a mass M. Two. So looking at the problem, we see that yeah, the marble block, The Marble Block, That's M. one has a mask equivalent to Mhm 559.1. So this is a marble block. This is a marble block. And then we have another marble block. Em too. Well, the second block, the second blog, sorry, the second block is a granite block and it has a mass M two. That's equivalent to 1 28.4 kilograms. So that's a granite block was seeing on the left. And then also the uh two blocks connected mm hmm with the rope that runs through a pulley system. So, so there's a pulley right here. This is up for me. And then what happens is that we're going to have a rope that runs around like this. I want to go back a little bit. So we have a rope that runs around like this. Uh and it's uh it connects these two. So these two are connected by a rope. Uh huh. Connected by a poorly that runs along like that. And the incline zone two angles, of course the fast one Is alpha and alpha happens to be uh 38 points three degrees. And then the second one is beta And beta happens to be equivalent to 57 0.2 degrees. Um And then also the other requirement is that we're going to have friction, I don't know the there's no friction on the boat. So there's no fiction on the road. We don't have any friction over there. But then block one deals with friction. So to block one, mm deals with friction. And The Coefficient of Friction for Block one. The coefficient of friction for black woman is equivalent to 0.13. And then the coefficient of friction for block too mm is the same as is the same as 0.31. The coefficient of friction we have for block one. and block too. And assume assume that coefficients coefficients of static and kinetic friction, yeah are equal. So coefficients of uh static and kinetic friction are equal. That's the assumption you're making. Mm hmm. Okay. And then the other thing is determined becoming the acceleration, determine the acceleration. And this is a mhm. Oh the model block. So that's the acceleration of this marble block assuming mhm determine the acceleration of the marble block, assuming uh positive ax is along the plane with someone. The plane that someone sits on. So what's going to happen is that this is our positive X. Direction. And uh the blog has a week and this way it happens to be really MG. So we want to be careful with the dimensions right here. So you know, this was this was our angle and this angle idea was alpha. So we just want to make sure that give it some space and this other animals beta. Mhm. And so I'll wait right here is M. G. Or rather M one G. M one G weight is M one G. Yeah. And then there is normal that uh poses the wait. There are two components to the wait. There's this one which is M G. M O N. G. Uh no sign of alpha. And then there's this other weight which is mong sign of alpha. And then there's a normal to the block that caused this way. This is anyone. And the same thing happens on this other side where we have um the second block we have mhm. M two G. And its components, we have two components. The first one is empty G. Call sign of beta. And then the second one is M. Two G. Sign off. Beat up. We want to have some space right there, mm hmm. Mtg sine of beta. Mhm And then this is a normal right here. This is into that's normal there's a tension on the rope that uh huh opposes, pulls the second block upwards. This way there's another tension pointing in that direction in the apple direction. And of course um this one so we're saying we have um the world flights of the for me. And uh and this friction over there. So we want to solve, we want to solve this particular problem. You know, we want to solve this problem right here. Mm So yeah some more forces will start by saying some of forces in the X. Direction is Uh huh. M. one a. x. So M. one G. Sign alpha minus detention minus the fiction F. One. Remember the friction is opposing the motion. So that means mm hmm if someone is falling downwards uh then the friction F. Is opposing the motion and this is the tension so fo and equals to uh M one A. So we could sell for the tension M one G sign of our farm minus one A minus. Remember the friction is a new one and one. So this is going to be me one and one. And then in the Y direction we have the summer forces in the Y direction is The same as M one A. Y. So N one minus mong co sign of alpha equals to zero. So we're saying this force and that force Uh the summation is zero because the system is in equilibrium. So we can say we can take, we can solve for N. One which is M. And G. Co sign of alpha. And then we'll plug in and one into that equation. So it ends up being mong sign of alpha minus and one a minus new one Mong kok sign of alpha. Mm And so for them to I would say this is for everyone, the mass of the the marble block. Uh this is the grand. And then the other one is the marble, this is the marble. So now we're focusing on the ground and was saying that um for the grand mm to some of forces in the X. Direction is M. Two A X. So we have the tension minus mtg sine of beta minus friction. The second fiction. So coming back to this, you can see uh you know, this friction have two is pointing in that direction and then we have the tension, so the tension is opposing the friction and this other force M two G B G silo beta this one, so they're moving in opposite direction and we also have something else happening. We can solve fatigued by saying this is M two A plus in two G sign beta plus F two mm And then some of forces in the wind direction is into a Y. So in the Y direction there's a equilibrium. And so we can solve for N two. And and to you could see for this friction right here we have t equals two M two A plus and two G beta was Youtube. And the normal to the friction. And so this normal to the friction. We can replace it into that equation and have a second equation. And to a plus two G sine, beta plus mewtwo, IM two G co sign of beta. Now we have two equations for attention and we're going to make them equal to each other. So mhm For this one was saying, and to a plus in two G sine of beta plus mewtwo and to the cool sine of beta, this is the same as the first one, which is M two G sign of alpha minus M one A minus new one. MRG course involved. Mm hmm. And so we're gonna combine couple of things were going to say AM one Plus 2. So we're combining this part of the equation and and this one into the into the left side and then plus M two G sine of beta um plus you to co signing beta. So yeah, yeah, what's happening right here is was saying M two G sign of data, that's this part plus M two G mhm You too whole sign of beta. If you look at mu to cause any beta, that's right here. So we're combining them mhm And then this is going to be equal to mong Sign of our farm and this new one Hussein of Alpha. So combining these two Equations into one equation. Uh huh. So these two equations combined into this equation and then these two equations yeah, combined into that equation. So that's what you have right there. And so right now we can solve for a by isolation so we're gonna have mhm G and one Gm one Sign off, a new one. Call sign off on the same too. Sign data plus me to co sign of data And then all over and one plus MP. And so the acceleration, it's going to be equivalent to 9.8, m/s squad. Uh 559.1 kg Sign of 38.3° -0.13. Call sign off mhm α, which is 38.3° then minus mm On 28.4 kg. That's the second was a sign of Bada, which is 57.2° since 57 plus 0.31. A court side of 57 joins two degrees And that's going to give us who want to divide that by some of the masses, which is 559.1 Killer runs plus 1 to 8.4 kg. So the acceleration Is approximately 2.28 m/s. Hope you enjoy the problem. Feel free to send any questions or comments and have a wonderful day of the wonderful

Previously, we've established that the coefficient of static friction on a slope determines the largest angle of that slope before the object will start sliding. Um And this equation, which has worked out in one of the example problems, is given given us the static friction needs to be greater than or equal to tangent of the angle. If the static friction is not greater than the tangent of the angle, then your object will slide down the slope. Um So we just need to plug in our value For the angle here, which is 24° And see if that plays out well. Tangent of 24° Is 0.44. And the coefficient of static friction that we are told in the problem is 0.63. So this proves to ourselves that this mass will not start sliding just yet. Part B of this problem asks us to figure out how much additional force given by a freezing body of water in the crack will be required to make this mass begin to move. And so we're gonna start by figuring out the forces that are acting on this thing. So in the y direction we have a portion of the wait force energy co signed data and that is balanced by the normal force. So normal force minus energy. Co ST peter, it's going to be equal to zero or in other words, normal force is equal to MG cosign data. Then in the parallel direction here we've got a couple of forces were going to have this force that we're trying to find. We are going to have a portion of the weight force in the parallel direction. And then we are going to have that friction force. The static friction force, well the static friction force and uh it's going to be equal to mute times the normal force. And we know that normal forces, Mg cosign theta. So this static friction force. His mu times energy co ST veda. And we want we are interested in the moments where this goes from being zero to actually having some acceleration. So we're going to figure out at what point does this exactly equal zero with the maximum coefficient of friction. Because that is the point when this will start to slide. And what is that force there? So we're just gonna plug this in here and I'm going to change colors to make it a little bit easier to see. So we're gonna have our mystery force MG sci fi to minus you. And then that's already factored out the MG. Um you cosign theta. Where does that equal? 0? Well, if we put this on the other side we get our force is equal to negative MG. So we'll just switch these around. We can just switch these around like so And so then we plug in our values here where M is 18 times 10 of the seven kg. And then we plug in The angle of 24°. And you should get a value this force again, using the maximum value of mu 63 You should get three Times 10 to the seven newton's.


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