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The figure below shows an initially stationary block of mass m on floor; force of magnitude F = 0.430-mg is then applied at upward angle @218(a) What is the magnitu...

Question

The figure below shows an initially stationary block of mass m on floor; force of magnitude F = 0.430-mg is then applied at upward angle @218(a) What is the magnitude of the acceleration of the block across the floor if (a) Us m/s2600 and Uk500?(b) What is the magnitude of the acceleration of the block across the floor if As 0.400 and Uk 0.300? m/s2

The figure below shows an initially stationary block of mass m on floor; force of magnitude F = 0.430-mg is then applied at upward angle @ 218 (a) What is the magnitude of the acceleration of the block across the floor if (a) Us m/s2 600 and Uk 500? (b) What is the magnitude of the acceleration of the block across the floor if As 0.400 and Uk 0.300? m/s2



Answers

Figure $6-20$ shows an initially stationary block of mass $m$ on a floor. A force of magnitude $0.500 \mathrm{mg}$ is then applied at upward angle $\theta=$ $20^{\circ}$. What is the magnitude of the acceleration of the block across the floor if the friction coefficients are and (b) $\mu_{s}=0.400$ and $\mu_{k}=0.300 ?$

So in this question we have the following situation, we have a block that's being pulled with the force F at an angle of data. So first let's complete our free body diagrams. We got normal force down, that's our first gravity down. Normal force up force of friction to the left. Now we also have to take this F at the angle and we got to break it up into its components are gonna get rid of this and break it up into how much of that is horizontal and how much of that is vertical. And we're going to do that is horizontal. There's gonna be f coast data and vertical is going to be f sign data and what we're gonna end up with is the following two expressions up and down. We're gonna be balanced. So G equals F. N plus at the signing peter and left and right. Let's assume that they're unbalanced. They don't necessarily have to be but at least if we start there you can always go backwards minus force of friction. Okay, so some of the things that we know, you know tha is 20 degrees. You know, force of friction is equal to 0.5. Plant and MGr force of friction. Just of course F. And we also have some formulas that we can use F. G equals M. G. Uh Net equals M. A. And force of friction static is less than or equal to us. And the normal force and force of kinetic friction is equal to UK comes from a force. Okay, so the first thing we have to do is let's determine our force of friction and our force or f coast data. Yeah, so F coast data, he's going to be equal to 0.5 MG and the co sign of 20 degrees. Okay, we plugged that into our calculator, Plugging in 9.8 for G. We get F costa Equals 46044 M. And yeah, so what we need is we need the force of static friction to be less than that in order for our block to be accelerating. So we have two different situations in part A us is 0.6 um UK is 0.4 That .4.5 mm. So let's look at our above situation and figure out the force of friction so we can move this top expression around and say that F. N. Is equal to F. G minus F sign here. So part of that means that our force of static friction is less than or equal to. We can force a friendship in new s times. Normal force, that's .600 times. You said normal forces, F G minus F santa. So it's M. G minus 0.5 MG times a sign. About 20 degrees. Yeah. Okay, so when we plug all this in, we get force of static friction is less than or equal to 4.87 M. Okay, so since the force of static friction is bigger than our F coast data, that means that F coast data is not overcoming static friction. So the acceleration has to be zero because it's not going to be moving it's not going to be able to overcome the static friction. Now if we go to part B, the only thing that's changed here Is that the coefficient of static friction is now only .4. So let's try this out. 2.4 times. M. G minus 0.5 MG. Uh huh. Can you sign of 20 degrees? Yeah. Yeah. And when we plug this in here. Mhm. Okay. We get force of static friction is less than or equal to 3.2496. M. Okay so this is good. Now we've overcome static friction so we gotta switch. So now let's calculate our acceleration using kinetic friction. So we'll start with our equation M. A. Equals F. G. Cost data. So 0.5 MG comes to co sign up 20 degrees minus our force of kinetic friction which in this case will be Our coalition of kinetic friction is only .3 times. Yeah MG -1 MG. Sign 20° and I ran out of room a little bit there at the end. Um But you get the idea so all of the ends actually go away which is really nice. So I am here and there and there and there. So plugging in 9.8 Fergie and solving you get acceleration is equal to 2.17 m per second squared. So in part A we did not overcome static friction, so the acceleration has to be zero because the block isn't going to move, it's going to remain stationary. In part B, it is going to move. So we have to switch over to kinetic friction. When we saw for acceleration, we get 2.17 m/s squared.

Now I can't actually see the diagram because I don't have a copy of the book but I think I can tell from the words what this is. So there's a block of math um on the floor we've got a force that's applied at an upward angle of 20 degrees. So on this block I'm going to draw this in red I guess on this block there's a force which I'm going to call P um of a magnitude of 0.500 MG. What does that mean? Mm Gee force of magnitude that MG That's not milligrams. I don't think I could it be gravity. Okay. Yes maybe yes. Times acceleration due to gravity. Okay. Got it. Um And this is data. I'm actually going to call it five though of 20 degrees. Okay. So in a the coefficient of static friction is 0.6 and the coefficient of kinetic friction is zero 0.5. All right. And what is the acceleration of the block? All right well I'm going to some the forces in the X. Direction that's going to be p times the co sign of fi and that's it. Oh no minus the friction force. I forgot to draw the friction force which is mu times the normal force which I'm going to call capital end. Now we also have a normal force and we have a gravitational force. Okay now back to what I was doing minus mu times the normal force. Um And erase. I just called this n not F. Seven. Just end. Okay. In order to figure out the normal force I need to sum the forces in the Y. Direction and they're going to be zero because the block is not lifting off of the ground. That's going to be the normal forces in the positive direction. And so is P co sign of five. Um Now and then in the opposite direction is the gravitational force which is M. G. Okay. No we don't know N um we know everything else here so just rearranging and is going to be MG minus P. Co. Sign five. So I'm going to a calculator does most calm and I'm going to declare my variables first. Bye equals 20 degrees. Um Oh we don't know. I am interesting interesting. We don't know him. Uh Do we know P. Yeah I don't really know P. Either. So I think I'm gonna not use the calculator. Let's go back to working on this. So what I'm going to do is I'm gonna substitute for N. Into the first equation. So that's going to equal P. Co sign. Fi minus mu times MG minus he. Co sign bye. Wait a minute. I don't think I should have a co sign in both of these. This second one here should be signed by P. In Y. Direction. Would be peace sign sign. Not peace sign. P. Sign. So that should be sign here. Okay so there we go PICO sine phi. That makes sense minus mu MG minus P. Sign five. Okay so just continuing with this PICO sine phi plus M. U. P. Sine phi. So I could actually factor out a P. Okay so now getting back to the question how much does the block accelerate? I know five. I know mu kind of dopey. Okay I see what I can do. I need to substitute in for P. Now I'm going to factor out MG. You know what I'm gonna know I'm going to factor out empty and then I'm going to distribute the 0.5 which is one half and then so I factored out the MG one half of that one half of that, the half does not go with this, so that's just going to be that. Okay um And that's going to be the mass times the acceleration. But notice moving upward there's really only one force acting in the X. Direction and that's that's P. Um And it needs to exceed the frictional force which will vary uh depending on whether P can overcome the frictional force. So the point is if I get an acceleration less than zero then I know it did not overcome the frictional force. So let's see what I can do in the calculator but wait a minute. Not yet, I'm going to cross out, I'm going to cross out the mess. The mass canceled out. Okay now let's see what I can do. So already put in five. Yeah. Okay. G. Well let's put in what G. Is 9.81 Um You it doesn't like meal. Um I was put in em from you and I'm going to use the coefficient of static friction which is 0.6 1st. I can get it typed in there. Yeah. Okay so now it's G. Times one half co sign of five minus mu over to sign. Yeah and keep having my fingers in the wrong spot. Bye minus mu. And that's it. So notice that I got a number that's less than zero. And therefore using the reasoning that I had since the forces. Again I'm going to explain the reasoning since the forces pulling in the positive direction. It has to go in the positive direction. If it gives me an answer that's negative. Well that means it did not overcome the frictional force. And so in part a the acceleration is zero. Did not overcome the frictional force. Let's go to be. What if M. U. Sub. S. Is 0.4 and um you sub K. Is 0.3. Well all I have to do is change my value of mu. Which I've listed as m here that gives me positive acceleration which means that the force was great enough to cause the block to move. But since it's moving now we need to switch to the coefficient of kinetic friction. Now I'm going to change it to three and it gives me 1.16 meters per second. Now it is not what's written in numerous. So there must be a mistake most likely. Although this isn't even problems so it's not 100% sure. Um But there must be a mistake in here somewhere so I'm gonna take some time and look over what I've done and see if I can find the mistake. And I've been looking here and I did find the mistake. It's right here. I have plus mu since data and then I just wrote minus musa science data by accident. So this needs to be plus new science data. This needs to be plus media science data. Okay? So now moving down, I'm sure that did not affect A but just to make sure we'll put 0.6 back in for mu trying to get the calculator here. So um you let's put this back at 0.6 and then this should have been plus right here. Yeah. Again we still end up with a being negative. So A does equal zero in part A in part B. Again we put in Emma's 0.4 and uh we get a positive number which means it exceeds the friction, the static friction force. So now we've got to go to kinetic friction. Please be the right number 2.17 which is the correct number. So erase this, this Okay. Yeah, so there we go. Uh and we get the acceleration is 2.17 m per second square. Thank you for watching.

All right. So in the first part of this problem, we've got an applied force of 12 newtons that is being applied at 25° above the horizontal on this five kg mass. And we want to know what the acceleration is. So, it's implied here. And we could prove it if we wanted to. But um we're going to prove it in part B of the problem anyway, so let's just go with it. It's implied that this mass does not lift up off the ground in this first part. So we're only going to concern ourselves with the X direction here. So the net force acting on the masses, mass times acceleration from Newton's second mom. And this is going to be equal to the X. Component of this force. So that's going to be F times the co sign of that angle. And this is on a frictionless plane. So the that is the only force acting on it in the X direction, in the horizontal direction. So this is our some of forces. So we can solve for acceleration by just dividing both sides by him. Get f. Cosign theta divided by M. And we'll get an acceleration of 2.18 m/s squared. Now in part B of this problem, we're supposed to figure out what force is required to lift this box off the ground. And so it's said that it's slowly increasing until that point happens. And so when what we're looking at here is the point in which the UAE forces are going to be equal or where the normal force in the line direction is going to dropped zero. So if you look at the forces in the wind direction um before we get to that point we won't have a net force. So the We're looking at the point where this stops being zero. So this is the M. A. In the Y. Direction. Um The forces acting in the Y. Direction are the weight force down the normal force up and the Y. Component of the applied force. So let's say that up is positive. So we're going to say the Y component of the Applied force have signed data plus the normal force minus the rate force. And we're looking for the point of which this normal force drops to zero at that point, that's when this box is going to lift off the ground. So at that exact moment um this will still hold true. And then if it gets any bigger than we'll start having an acceleration here, so when this goes to zero, F. Science data is going to equal MG. So this is the point that we are looking for. Um So when that happens we just solve this for F. We get an F. Is equal to MG over signed data. And we get a applied force of 116 newton's forget 366 So when this force becomes 100 and 60 newtons, that normal force will drop to zero and our box will start to lift off the ground. Now the final part of this problem part C asks us what the acceleration will be at that point. What? We can just take this force and shove it in here into this equation and that will give us our new acceleration. So let's just do that. So our equation is this still have cosign Theta over temp. And if we plug those values in 1 16 times co sign of 25° divided by five, you end up with 21.0 meters per second squared

So for part A to find the for the magnitude of the kinetic frictional force, we can apply Newton's second law in the Why Directions. So the force normal minus f sign of Fada minus M g. This is gonna be equaling 20 So we can say that. Then the forced normal is gonna be equaling two. This would be 15 Newton's multiplied by sine of 40 degrees. This would be plus 3.5 kilograms multiplied by 9.8 meters. Chris second squared and we find that the normal forces equaling 44 Newtons. So we can say that the kinetic frictional force would be equaling to the coefficient of kinetic friction times the normal force. And this is equaling 0.25 multiplied by 44 Newtons giving us 11 Newtons for our answer for part A For part B, we're going to apply Newton's second law to the X axis. We're here f co sign of theta minus. The kinetic frictional force is gonna be equal to the mass times acceleration. So the acceleration would be equaling 2 15 Newtons co sign of 40 degrees. This would be minus 11 Newton's and then this will be divided by the mass of 3.5 kilograms. The acceleration is found to be 0.14 meters per second squared. This would be our final answer for party. The fact that this is positive means that the acceleration is in the positive X direction. That is the end of the solution. Thank you for watching.


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