a-box-with-a-mass-of-10-mathrmkg-is-placed-on-an-inclined-plane-that-makes-a-60circ-angle-with-the-2

Question

A box with a mass of 10 $\mathrm{kg}$ is placed on an inclined plane that makes a $60^{\circ}$ angle with the horizontal. The coefficient of static fraction $\left(\mu_{\mathrm{s}}\right)$ between the box and the inclined plane is $0.2 .$ What force must be 
(A) 32.7 $\mathrm{N}$
(B) 48.3 $\mathrm{N}$
(C) 56.2 $\mathrm{N}$
(D) 76.6 $\mathrm{N}$
applied on the box in order to prevent the box from sliding down the
inclined plane?

Salamat Ali · Accepted Answer

their forces on a box on the inclined plane are following. So we have a box here, the force downward. He&#x27;s if wait, scientific data scientist that where the force of friction is outward. So force of friction here and the normal force in this direction and downward. He&#x27;s the the compliment off the wait. So where it is here? The compliment of right here is if the beautiful wait cost that I see then equating the forces on X axis and why direction? Access. Let&#x27;s say this use of our X and this is why So why? Then we can write our forces. Someone forced on the box is equal to zero. We can write. It&#x27;s f normal is a call to if normal is equal to force. Wait times course later course, Vicar. Um then we can write a forces on why Direction? Which our effort Clyde will be cool too. N g scientific minus the forces of friction signs minus a mule mg n g cost data course later, that&#x27;s obstructing the values. Um, we have a mass is a 10 traditional exertions. A 10 sign off 60 sign off 60 year and minus use little 0.2 when deployed by the Mass will be over 10 times 10. I didn&#x27;t times course or for 60 of course. Over. A six see 60. Then they applied force. We get by simplifying this expression is 76.6 Newton. Then, looking into our options, we see Option D is the right answer is the right answer.

Kristela Garcia · Answer

in this problem, we&#x27;re given a box sitting on an incline plane of an angle of 30 degrees, and we want to calculate the coefficient of static friction between the box of the incline plane that must exist in order to get this box at rest. So drawing everybody diagram. We have a normal force perpendicular to our surface force of gravity street down. And we have a force of friction going up the ramp to hold this object in place. And if we look at our components of gravity, we have the component that&#x27;s perpendicular to art inclined and a component that&#x27;s parallel to our incline. So this component, this, um, adjacent side will be m g co signed data, and this opposite side will be mg signed data. And because object is at rest, we can say M G signed data must be equal to my friction. Now we want to calculate the coefficient of Max static friction in order for this to be held at rest. And so my maximum assuming this 30 degrees is my maximum is going to be my coefficient of static friction. How&#x27;s my normal force, which is mg co signed data in this case. So we have new times, mg coastline data, my MGs cancel out, and I&#x27;m left with signe data divided by co signed data, which is a trick function of tension data gives me my coefficient of static friction so I can do 10 geant of 30 degrees to find my coefficient. Um, 0.58

Ryan Hood · Answer

let&#x27;s start out by drawing the force diagram where the forces directed down the ramp. And so, if this is my object on the ramp, I&#x27;m a normal force going this way. Gravity going down. I have the applied force going this way. I have a friction force opposing the motion, and I&#x27;m going to break up this gravity into forces along and perpendicular to the ramp. And so this is the angle that they give us. And so this component here is mg time sign of 55 which is the angle they give us. And this is M. G co sign of 55. And now we&#x27;re ready to applying Newton&#x27;s second law. This is my co ordinate system, By the way, it&#x27;s positive X goes down the ramp and positively goes upwards from the ramp. So some of the forces in the white direction equals zero. Since six sellers in the White Direction zero, this implies that the normal force is equal to N g co sign of 55. Since those two forces air, the only force is in the white direction now, uh, blinking in values. Here I get that the normal force is 56.21 Nunes. And then when I played this in to find the friction force I get, that&#x27;s equal to the coefficient of kinetic friction, since the box is moving times normal force, which I should be consistent and indicated by a lower. And as I did here and then when I played the sand, I get 16.86 Nunes. Now I&#x27;m going to apply Newton&#x27;s second law on the extraction. In this case, there is an acceleration. So it&#x27;s NyQuil zeros he called a master acceleration, plugging in all the forces in the extraction. I get positive. Nicholas, positive exponent of the way, grabbed, you should say, minus the friction force, because it&#x27;s a new direction and Siegel amassed an assertion. The X we&#x27;re solving for the acceleration. We just found the fresh and force the applied force is given and the masses given. So everything in this equation is given and we&#x27;re solving for acceleration. And so when we put all that into the calculator, we get that acceleration is positive. 18.34 meters per second squared. So this is acceleration down the ramp that the box experiences in part be the only thing that changes is now the applied forces upwards and the friction forces downwards. So they swap places. The normal force is still this way. Gravity is still down, and it still has the components, the same exact components. I won&#x27;t write out again, but they&#x27;re there now. The friction force is still 16.86 news because it&#x27;s derived purely from the normal force and the coefficient of connect friction. And both those qualities did not change with this new situation here. So this is still the friction force. And now I&#x27;m just going to a straight to applying Newton&#x27;s second law on the extraction so that there is there and I just need to substitute in for my forces. I&#x27;ve a positive aversion force now, a positive way, contribution or gravity contribution, said Sanford Fire, and I subtract my applied force because it&#x27;s now in the negative X direction, and this is equal to mass times acceleration. The X now once again, like last time, everything here is no one except for the acceleration. And that&#x27;s what we&#x27;re solving. Four. So when I do self word I get, the acceleration is equal to negative 2.286 meters per second squared. So the acceleration is up the ramp by this amount and so much less because gravity is working against us now and this

Aaron Shoolroy · Answer

right, Jeff, before problem 23 we got two blocks and side by side. We&#x27;re pushing with 100 Newtons on the bigger block on the left hand side. And they want to know what is the force that the smaller block pushes on the bigger block with. And the key here is gonna be drawing correct free body diagram. So let&#x27;s start with the 20 kilogram block. That dot represents the 20 kilogram block, the forces acting on it, our M G going down and the normal force going back up that blocks not accelerating vertically. So those air equal, we&#x27;re pushing on that block with 100 Newtons of force to the right block. The five kilogram block is pushing back to the left. And let&#x27;s call that the force from Block. Oh, I can&#x27;t remember what they named these guys. I believe this is a and this is B. So let&#x27;s call the force on A from B right now. The key to understand this problem is a free body diagram also for what&#x27;s going on with the five kilogram block. Okay, because I know there&#x27;s a Newton&#x27;s third law pair here, so yeah, we&#x27;ve got the weight going down here. We have a normal force growing up here on. Then we&#x27;re gonna have the force from be on a Let me draw that a little bit better, because I know that this is an action reaction pair. So the force of be on Okay, uh, so let&#x27;s see here, um, those forces were going opposite, so Force A B is equal to force B A. And if we&#x27;re gonna find one of them, we can find either one because the magnitude of the force is gonna be the same. So looking at Newton&#x27;s second law over here for Block B, the Net force acting on Block B in the extraction is equal to the mass of block B times its acceleration in the extraction, the only horizontal force acting is that Force B A. And that&#x27;s gonna equal the mass of Block B, which was given us five kilograms times its acceleration. Now we know the mass, but unfortunately, we don&#x27;t know the acceleration of either block. I know that they&#x27;re gonna have the same acceleration, so we actually have to find the acceleration first. So I&#x27;m gonna use a trick since both blocks are accelerating at the same rate. I&#x27;m gonna look at the net force acting on the system as a whole. That&#x27;s gonna equal the mass of the system as a whole times that acceleration. So we are talking about forces acting in the extraction because that&#x27;s the direction of the acceleration. So let&#x27;s be more specific there. So let&#x27;s talk about the forces acting in the extraction. We have this force a B that&#x27;s going back to the left negative F A B. We have the 100 Newton force going to the right, so it&#x27;s positive 100 and we have forced be a that&#x27;s going to the right. So that&#x27;s also positive. And that&#x27;s going to equal the total mass of the system, which is 25 times the acceleration. Okay, now, here&#x27;s the trick. F A V is equal to F B A. One of a negative, one of them positive. They&#x27;re gone. They cancel each other out. Internal Forces thes Newton&#x27;s third law action reaction pairs will always cancel out when we consider the Net force acting on the system because they&#x27;re always equal and they&#x27;re always in opposite directions, so that leaves us just with 100 Newtons of force acting on the system, divided by the total mass of the system. Then that&#x27;s the acceleration four meters per second squared case of four meters per second squared. That means that Block B is accelerating to the right at four meters per second squared. They&#x27;re going back up here. The force of be on A or A on B is gonna be the massive block B times its acceleration. That would be 20 Newton&#x27;s. Okay, so if it&#x27;s 20 Newton&#x27;s here. Then it&#x27;s gonna be 20. Newton&#x27;s here. Okay, so it&#x27;s worth noting. Why does block a accelerate to the right? Because it&#x27;s got a net force acting to the right, hundreds of the right, 20 minutes to the left. Why does block B accelerate to the right? It&#x27;s just got horizontal force of 20. Newton&#x27;s acting on it to the right. Okay, so those equal and opposite action reaction pairs will always be equal and opposite. All right. If I pushed with 100 board sorry harder than 100 newtons of force, the amount of that equal and opposite force could change. It could rise to 30 40 50 Newtons, but it will always be equal and opposite. All right, the magnitudes will be the same, pointing in opposite directions. That&#x27;s that problem done and dusted.

Carolyn Shasha · Answer

this is question. Five points of me too. We have a six kilogram box sitting on a ramp that&#x27;s inclined at thirty seven degrees. The coefficient of kinetic friction between the box on the ramp is point three and we are told that a horizontal force is applied and the box is moving up the incline with a constant acceleration of three point six meters per second squared. So I&#x27;ve drawn the free body diagram. Here we have this horizontal force being applied. The ramp you can picture is here, so friction is pulling is opposing the motion. So the friction is going to be pointed down towards the ramp. We have the weight of gravity pulling straight down, and then the normal force against the ramp will be, um, perpendicular to the ramp. This is R and normal force. So let&#x27;s first consider the forces that we have here. We have a force of friction. So, uh, let&#x27;s explicitly right out that our friction force is going to be its kinetic friction since the box is moving. So it&#x27;s going to be our coalition of kinetic friction times the normal force now the normal force here. Since we have two forces that are pushing on this block that are ah perpendicular here or that are going to cause this force perpendicular to the ramp. We have to consider the normal force due to each of these components separately. So the normal force is going to have one component from gravity from this Ah, great force. And that&#x27;s going to be prem g times the co sign of data. Here we break this up into components perpendicular and parallel to the ramp, and then we&#x27;re also going to have the component from the this force and for you the same thing we&#x27;re going to get that the normal force from the horizontal forests is f times signe of data. So now we can find our total friction forests, Clyde just plugging in to our normal for us right here. So that&#x27;s our friction force. So now we want to set up our Newton&#x27;s second law creation, so we know that the sum of the forces is going to equal mass times acceleration. So let&#x27;s think about all of the forces, um, going up the ramp here since we know that the acceleration is up the ramp. So we&#x27;re going to consider this and as the positive ex direction. So we&#x27;re gonna have that component of the forest Will be F Times Co sign of data. Me, we&#x27;LL have a co signer Vada That&#x27;s gonna be the component of the forest that&#x27;s pushing up the ramp. Then we&#x27;re gonna have our friction force, which is proposing that some linus are friction forces that&#x27;s pulling it ground. And we&#x27;re also gonna have, ah, gravity pulling down. So the component along the ramp there going to be m g times sign Or expect that though this is going to be our total force acting on this No, in the direction along the ramp and that&#x27;s equal to mass times acceleration. So now we want to solve this for F So remember, we have I&#x27;m goingto at this point plug in our whole expression for our friction for us since they there&#x27;s an expression for ah, since we cf show up there and let&#x27;s to simplify this. Let&#x27;s first distribute out this factor of Mu K. So that&#x27;s going to give us a minus. You can front here on, and now we can group the like terms together. So we want to get F on its own. So let&#x27;s ah, move all the other terms to the right side. We&#x27;re gonna add this to the right side, and I&#x27;m going to add this term to the right side. So we now have these two terms with F so I can pull out the factor of F. That&#x27;s goingto leave co sign of data minus mu K sign of data. And so, finally to solve for f B. Just wanted provide both sides by this entire term here. So this looks like pretty messy expression. But all of these are something our variables that we know. So we know that m iss six We know the acceleration here is three point six uk is point three, gee is nine point eight and data is thirty seven degrees. So we can just plug in everything directly here. And when we do that, we should find ref equal to one hundred and fifteen. Eunice

Jacob Paiste · Answer

Alrighty. So we have this five kilogram box that&#x27;s sitting on a horizontal surface. Here it is. M equals 5.0 kilograms. And there&#x27;s a coefficient of kinetic friction on this surface, and that&#x27;s equal to 0.50 And the horrors that we there&#x27;s a horizontal force that pulls the box here and that horizontal force. It&#x27;s not given to us. We want to. And keep in mind, though, that this is, um, not given to us, and it&#x27;s gonna keep it the equal sign that we have to know. One thing at the velocity when this thing is in motion is equal to a constant. So the net Force is gonna be equal to zero. We need to find the work done by a few things. Euro Last thing. We should definitely also mention the same gonna move a distance here. Distance D We may make some more space here. My coefficient of static friction. I can move this over here and take my displacement vector back here. Cool. Oh, there we go. And back to my marker Heikal so d equals they give it to us and centimeters. So let&#x27;s keep it like that for now 10 centimeters. Let&#x27;s figure out the work done by the supplied horizontal force by the Frictional Force and by the Net Force. So three different things we got to figure out here. So let&#x27;s try to figure out the one that work done by Ah, my friction. Right. Do that first. So we know that the work done by friction is going to equal. It&#x27;s always negative here because that frictional force is gonna be opposite of our motion. I won&#x27;t put a negative in. Yeah, logistician you guys what I mean here. So number frictional force is equal to the normal force multiplied by the coefficient of friction. Normal force here is gonna be equal to N G and the coefficient of friction music camp and removing a distance d just 10 centimeters. Now remember, this is the force multiplied by the distance and then multiplied by the angle between the vectors. That&#x27;s your normal work equation that we use. This guy right here is negative one. And as results, you papa negative one of everything here. So have a negative M g m u sub k times D, which is equal to a negative five times 9.8. Not gonna bother my units here. That&#x27;s the gravity I pointed. One multiplied by the coefficient of friction. And for the 10 centimeters, seven multiplied by 70.1. And then we will get the work done by friction. And that is equal to negative. 2.45 Excuse me? My marker had his little freak out there. Jules. All right, that&#x27;s the work done by friction. All right? No. Remember, this thing is moving at a constant velocity. All right? So that means the work total has to be equal to zero, and work total is going to be equal to your work off your friction multiplier. Um, and then plus the work done by the, um, Applied Force. And it&#x27;s equal to zero because work totals equal to zero. What does this mean? This means that these guys are equal to an equal in magnitude opposite in sign. So you work done by your applied force is going to now be a positive 2.45 Jules

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Given a solution of insulin (molecular weight $=5…

Problem 11 Hard Difficulty

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