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Block (mass kg) moving 20 dcgrccs inclinc surtace Cocfficicnt of kinctic friction betwccn block and the surface This block is connccted block (mass kg) by massless ...

Question

Block (mass kg) moving 20 dcgrccs inclinc surtace Cocfficicnt of kinctic friction betwccn block and the surface This block is connccted block (mass kg) by massless cord that passcs over massless and frictionless pulley: Draw free-bods diagram:Find the acceleration of each block and the tension in the cord using the Lab simulation Videol by applying Newton " Law of motion: Show stepsTwo objects (40.0 and 25.0 kg) are connected by massicss string that passes over massless frictionless pulley:

Block (mass kg) moving 20 dcgrccs inclinc surtace Cocfficicnt of kinctic friction betwccn block and the surface This block is connccted block (mass kg) by massless cord that passcs over massless and frictionless pulley: Draw free-bods diagram: Find the acceleration of each block and the tension in the cord using the Lab simulation Videol by applying Newton " Law of motion: Show steps Two objects (40.0 and 25.0 kg) are connected by massicss string that passes over massless frictionless pulley: The pulley hangs from the ceiling_ Use - free-body' diagram Find (4) the acceleration of the objects and (bZ the tension in the string Show steps



Answers

Block $A$ in Fig. $6-56$ has mass $m_{4}=4.0 \mathrm{kg}$ and block $B$ has mass $m_{B}=2.0$ kg. The coefficient of kinetic friction between block $B$ and the horizontal plane is $\mu_{k}=0.50 .$ The inclined plane is frictionless and at angle $\theta=30^{\circ} .$ The pulley serves only to change the direction of the cord connecting the blocks. The cord has negligible mass. Find (a) the tension in the cord and (b) the magnitude of the acceleration of the blocks.

For this problem on the topic of false and motion. We are shown in the figure two blocks A and B with masses four kg and two kg respectively. And we're told that the coefficient of kinetic friction between B and the horizontal plane is 0.5. The implant plane makes an angle of 30 degrees to the horizontal and the fully changes the direction of the card connecting the blocks. We want to find the attention in the cord as well as the magnitude of the acceleration of the two blocks. Now, the free body diagrams for A and B are shown as we have here and Newton's second law gives us for block A. M. A. G. Science theater, which is the component of the weight down that plane minus detention T. Is equal to its mass. M. A. Times the acceleration A. Now for Block B we have the attention T minus the frictional force F. K. Which is due to kinetic friction is equal to M. B. Times its acceleration. A acceleration is common for both blocks. The frictional force F K. Is equal to um UK times the normal force FNB acting on block B, which is um UK. The coefficient of static friction times the weight since they wait and normal force balance. So that's M. B. Times G. That was the U. K. And so bye combining the above equations. We can solve for the tension. We get the tension in the string T. To be M. A. Times M. B. Of uh M A. Plus MB into Sign theater plus um UK times G. These values are all known. So if we put them in we get full kg times two kg divided by for plus two kg into the sine of the angle of incline 30 degrees. That's the coefficient of kinetic friction. 0.5 times 9.8 m per square second, which gives us detention in the cord to be 13 newton's. Now that we have the tension, we can calculate the acceleration and so the acceleration A. Is equal to M A. Sine theta minus um, UK m b over M A plus MB times G. And so again, we have all these values. This is for K G times the sine of 30 degrees minus 0.5 times two kg. We're suppressing the units, again divided by four kg. That's two kg. All of this multiplied by the acceleration due to gravity 9.8 m per square second. This gives us the acceleration of the system to be 1.6 meters per square second.

In this exercise, we have the system that shown here in the figure of the left that consists of two blocks of mass and one equals 2 kg and and to equal to 6 kg that are connected by, ah, string that passes through a pulley that has a mass of 10 kg and a radius of zero points. 25 m. The second block and two is on an inclined plane that makes an angle off 30 degrees with the horizontal. And both blocks are subject to friction. And because the friction coefficient new is equal to 0.36 and I go first of all is to draw the free body diagram for both blocks and the police. So for the first block in one, we have him the bucket one subject to the force of gravity that's equal to him one g a normal force that I'm just gonna told man in. Then we have the tension floors pointing to the right, and we also have the frictional force which is just gonna call laugh, pointing to ah to the left notice that F is equal to the normal force times mu and the normal forces just in one g. So having one g mute, this is a frictional force for the second block. We have the gravitational force that can be decomposed into two components, one that is perpendicular to the surface and has a value, as we know from inclined plane studies that this is mg times the cool sign of theater. And this here's sorry, this here's data as the force Ah, parallel to the surface, Which is it which would end one g sign of data. We also have attention force actually, let me just rather normal force first. Then we have the tension floors t two. That's attention that the string exerts over the block and we also have the original force f. And I'm just gonna call it f two and I'm called the first frictional force from the first look at floor. So f two is equal to the normal, which is in I Geico sine theta times mu. And finally, for the Cooley we have the gravitational force and a normal force g there. Oh, no one just gonna call in. And we also have the attention to you on that acts, uh, to the left on the Pooley and attention T two that acts to the rat every it acts downwards and to the right. Okay, so these are the free body diagrams for the three object and in question A. We have to determine the exploration off the to blocks. So let's move on Depression A and we're gonna use the these three body diagrams. So, first of all, I'm gonna write an equation for the acceleration of the first block. Notice that ah, horizontally that the block external accelerates horizontally and the horizontal force is equal to t minus F one. That's the net horizontal force or have t minus F one is equal to m one a. Then actually it should be t one. This is the Yeah, you should have just put a one here that this is the attention of the string upon the first block. Then for the second block, we have that along prophetic. I'm sorry, Parlow. To the the surface that is parallel to the exploration we have that the total forces in one g times the call sign. Okay, the science data minus the official force F two minus teach you and this is able to him two times the acceleration a notice that the exploration of both blocks is the same. And for the Pooley, instead of writing a force equation for the center mass, I'm gonna write a torque equation. Notice that, um, the Force t two causes the the pulley to rotate clockwise. So the torque we were positive, so have to t two times are the force t one causes the fully to rotate counterclockwise. So it's negative. Mind at the torque in negative minus t one are. And this is equal to the moment of inertia, of the holy times, the linear acceleration. So we have these three leaves. Okay, I'm gonna write them a little differently because I'm gonna, uh, substitute everything that I can as explicitly as I can to have tea one minus f one. And they said before F one is just and one gene you and this is able to m one a. Then we have and one G science data minus f to which is and one Geico. Science data sounds mu minus t two, and this is equal to him to a then we have de to Manistee one are and this is equal to the moment of inertia, of the ability which is n r squared over two. That's the moment of inertia off a cylinder times the the angular acceleration Alfa, which is the leader Acceleration A divided by the radius of the pool. Notice that the radius here castles out. So we're left only with two minus. T one is equal to m A over to. So this equation to this week in here I'm going to give the name one to this. I'm gonna give meaning to to this one. Here, the name three. Then what I'm gonna do is to some the three equations one plus two plus three. So, miss, um, through them, and notice that the teeth cancel out because we have 21 minus C, 2% u minus 21 So we only have to care about the other terms. So I have minus and one g nu plus M one g, scientific data minor applied. This should be in a two year I team two because this is the mass of the second block, not the first one. I just carried out a mistake, but I should be in too. So this here should be him too minus and to times g times a cool sign of favorite times mu iss and this is equal to and one pleasant to less came over to a so a times in one pleasant to plus an over to is equal to G times am two times side of data minus mu time School side of data because ser minus and one mu so a is equal to g times and to science data minus mu Cool science data Klein is in one you divided by and one wasn't too. Plus I am over to and now we can substitute the numbers explicitly. So ji is not 8 m per second square times and to which is 6 kg. So it is a sign of 30. Mine is your 0.36 times the co sign. Ah, 30. Mine is in one which is 2 kg. Times is your 0.36 divided by 6 kg was 2 kg plus 10 kg of I device to. So if we carry out the copulation, A is equal to 0.31 m per second square. So this is the exploration of that. Your blocks, Any question be we have to calculate the tensions in the two sides of the of the string. So t wanted to to s. So basically, what I'm gonna do is to go back to the equations that I wrote before. So I'm going to use first the first ingredient in order to find t one. So we have we just just copy The clay didn't up here. So is he. What? Minus him Once she knew is equal Chem one A. So T Wat is able to end one A because G mute. So this is 2 kg times a which is, uh, 1 31 m per second square plus g which is not what, 8 m per second squared times you're going 36. So it's here. One is able to 7.68 newness and then t to in order to fight you two just gonna use Equation three, which says that t two minutes to one in which we may over to so two to minus t. One is a royal A over to so teach two is equal to t one plus and a or two no t one, as we found, is people to 7.68 Newtons plus an, which is uh, 10 kg divided by two times is your point 31 m per second squared. So t two is able to 9.23 Newtons. This is the value off the attention of the second part of the stream and they include enough over question.

For this problem, On the topic of rotation, we have two blocks of mass two and six kg, respectively, connected by massless string over a pulley in the shape of a disc. The pulley has a radius of 0.25 m and a mass of 10 kg. Now the blocks are allowed to move on a fixed block wedge of angle 30°,, as we can see in the figure, The coefficient of kinetic friction for both blocks. 0.36. We first want to draw a free body diagrams of the blocks and pulley and then determine the acceleration of the two blocks, as well as the tensions in the string on both sides of the pulley. Now, we've drawn the free body diagrams for mass one, the pulley and mass. To. Now from these We look at em 1 1st and the some of the vertical forces must equal I am a Y. By Newton's second law and so On. M. one. We have the normal force in one minus the weight of mass one, which is M one G Is equal to zero. And so in one is equal to M one G, which is 19.6 Newtons. And the frictional force Acting on Block one FK. 1 is equal to the coefficient of kinetic friction in UK times in one Which is 7.06 newtons. And if we take the some the horizontal forces or the forces along the X direction. Again, this was equal to M. A. X. from Newton's 2nd law. And we get minus seven point 06 Newtons Plus the tension T. one is equal to the mass two kgs times the acceleration. A. So the acceleration is purely horizontal for block one. And we call this equation one, there's two unknown zia. Now, for the pulley we have the sum of talks must equal to I alpha. The angular form of Newton's second law and so minus T. One times are bless T two times are must equal half M R squared, which is the moment of inertia II times. The angular acceleration is acceleration A divided by our. So from here we get minus T. One Plus T two is equal to a half times five kg times acceleration. A And so we call this equation too. Now for mess em too, we have the normal force into minus M two G. Co sign of theater is equal to zero and so we can rearrange and we can solve into so we get into is equal to six kg Times G, which is 9.8 m/km2 times the call sign of 30 degrees, Which gives us the normal force acting on block two To be 50.9 Newton's and the frictional force acting on block two is F. K two, which is the coefficient of kinetic friction mieux que times in two. And this is equal to 18.3 Newton's. And so we can see that -18.3. Newtons minus T. Two. Bless em too. Science data is equal to M to A. So Newton's second law for block too. So we get -18 0.3. Newtons minus T two plus 29.4. Newton's is equal to six kgs times acceleration. A And we'll call this equation three. So now we have the three equations with three unknowns T one, T two and A. So firstly we want to find the acceleration and we can add The three Equations 1, two and three. And if we add these three equations we get minus 7.06. Newtons minus 18 0.3 Newtons bless 29.4 Newton's is equal to 13 kg times A. So rearranging, we find the acceleration A to be 0.309 meters per square second. Now that we have the acceleration, we can find the tensions. And so we have T 1 to be two kg Time 0.309 m/km2 Plus 7.06 Newtons, Which gives us T one To be 7.67 Newton's and T two. He is equal to seven .67 Newtons plus five kg Times acceleration 0.309 meters per square second, which gives us T to to be nine .22 Newtons.

So here we control all the system. It's is this a frictionless massless pulley? Aah! This would be, ah, maths. A going up would be the force normal of a going down would be, of course, um sub a g and going to the right butts. This would be massive. Be going to the right here would be forced. Tension going up here would be forced. Tension going down would be m sabi tree and this is the full free body diagram for the system. Now we can say that the force tension is going to be equal to rather this would actually my apologies. This would be your answer for part A. So just party just asked us to draw the free body diagram. This would be the full free body diagram for part B. However, now we need to solve for the acceleration. So here. Ah, for block A. There is no motion in the vertical direction. Therefore forced normal of a wood equal the mass of a g again. This plane right here. This table is perfectly horizontal. Therefore, the normal force of block a on block a rather would be equal to the weight of block A We can apply Newton's second long to block a in the ex direction. This would equal force tension and this would equal the mass sub a times acceleration of a in the X direction. We can say that the sum of forces for block B in the UAE direction would be equal to M C B G minus forced Sebti Aah! This would equal and sub B times a. So be in the wind direction. Now the two blocks air connected. Therefore, the acceleration of block A in the ex direction with the equal to the acceleration of block B in the wind direction. And we're going to call this simply a So we're gonna lose the sub scripts for a and then we can combine the two force equations. We can essentially combine these two equations to solve for a so we can say that forced tension would then be equal to m sub a time I'm so be she with minus force sub t would be equal to m sub b a and then we're going to say that m c b g minus m c a a cool m sub b times a uh this would become massive eight times today plus massive B times A would equal mass Sabi G And we can say that the acceleration would then be equal to the massive be times G divided by the sum of the masses. And this would be your answer for the acceleration. At this point, we want to find force tension. We know forced tension would be equal to again m sub a times, eh? Therefore, this would be equal to g times m sub a m sabi divided by again the sum of the masses m sub A plus I'm sabi So this would be your force tension here That is the end of the solution. Thank you for watching.


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