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Draw free body diagram (FBD) showing all of the individual fores the system consisting of boxes acting and together. Indicate the directions of the five diagram. If...

Question

Draw free body diagram (FBD) showing all of the individual fores the system consisting of boxes acting and together. Indicate the directions of the five diagram. If a force doxes nOt fonee bclong - Jif the listed below; choosing from vcctors A-D in the on the FBD. enter N.Eg- tirst forcc the system. enter ANNNN. direction _ rmaining forces arcnnt applied t0box pushing eround the upplicd force , pushing block 3 gravity pulling on box box pushing on box Dox pushing boxAnswer: IncorrectSubmitAnwens

Draw free body diagram (FBD) showing all of the individual fores the system consisting of boxes acting and together. Indicate the directions of the five diagram. If a force doxes nOt fonee bclong - Jif the listed below; choosing from vcctors A-D in the on the FBD. enter N.Eg- tirst forcc the system. enter ANNNN. direction _ rmaining forces arcnnt applied t0 box pushing eround the upplicd force , pushing block 3 gravity pulling on box box pushing on box Dox pushing box Answer: Incorrect Submit Anwens Last Answer: 6,ON trica 7/15 _ 3 . [Ipt] Find the contact forcc between boxes anld Answer: Supmm AIl Answere Last Answer: 2 [N Kot correct Ericb 4/15 Hint: You will need draw two more FBDs. and IL. Choose box as your object of inlerest for FBD [ and apply Newton > Endlt Choose all three boxes VOUI object of interest for FBD II. The advantage of doing this is that there are no unknown forces for FBD I. You then solve for the acceleration which will help you solve the equation you obtained from FBD [Ipt] Find the contact force betwcen boxes and Answer: Subrl Answlone



Answers

When a force of $500 \mathrm{~N}$ pushes on a 25 -kg box as shown in $\underline{\text { Fig. } 3}-$ 18, the resulting acceleration of the box up the incline is $0.75 \mathrm{~m} / \mathrm{s}^{2}$. Compute the coefficient of kinetic friction between the box and the incline.
The acting forces and their components are shown in Fig. $3-18$. Notice how the $x$ - and $y$ -axes are taken. Since the box moves up the incline, the friction force (which always acts to retard the motion) is directed down the incline. First find $F_{f}$ by writing $\sum F_{x}=m a_{x} .$ From $\underline{\text { Fig. } 3-18}$, using $\sin 40^{\circ}=$ $0.643$, $$ \text { + } \sum F_{x}=383 \mathrm{~N}-F_{\mathrm{f}}-(0.643)(25)(9.81) \mathrm{N}=(25 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) $$ from which $F_{f}=206.6 \mathrm{~N}$. We also need $F_{N}$. Writing $\sum F_{y}=m a_{y}=0$, and using $\cos 40^{\circ}=$ $0.766$, $\left[\Sigma F_{y}=F_{v}-321,4 \mathrm{~N}-(0.7660)(25)(9.81) \mathrm{N}=0 \quad\right.$ or $\quad F_{v}=509.3 \mathrm{~N}$ $$ \text { Then } \mu_{k}=\frac{F_{i}}{F_{v}}=\frac{206.6}{509.3}=0.41 $$

Hello. And this problem we have a box that weighs 70 kg And this box is being pulled with a force of 400 newtons at an angle of 30°. So here's the free body diagram as we can see And that's our 400 force Acting at an angle that is 30° from the horizontal. So the question here is to find the exploration of this box here on the floor, given that the coefficient of kinetic friction between the box and the floor is 0.5. So from here we know that we need first to indicate how much is the frictional force. And to get the frictional force we need to determine the normal force. So here of course we have the normal force acting upward and of course the weight of the box itself is acting downwards. So here that's the force of weight and here that's the normal force. And this as we said, is the frictional force. So let's start by applying the equilibrium or the balance of forces in the y direction. The sum of forces in the Y direction should be equal to zero because there is equilibrium in that direction. So from here, if we take upward to be positive, so the force of normal minus the force of weight. But here we also need to include the component or the vertical component of this. 400 notice force Because this 400 Newtons force have a vertical component and a horizontal component in this direction. So here we also have with a positive sign. 400 Noten's And here to get the vertical, we multiply by sign the angle All of this is equal to zero. So from here the normal force is equal to the force of weight, which is 70, Multiplied by 9.8. 1 minus 400 sign 30 signed 30 is equal to half. So here we have 200 And calculating this, we know that the normal force now is equal to 400 86.7. Noten's. So that's the first step and that's from the equilibrium in the vertical direction. Now we need to get the exploration in the X direction and we know here that the sum of forces in the X direction should be equal to the mass of the box, multiplied by the acceleration in the X direction. So here, let's take a look again at this free body diagram. We have the horizontal component of this. 400 newtons acting to the right while the frictional force is pulling or acting in the opposite direction. So here the summer forces is equal to the force. Or let's write it directly here We have 400 And this time multiplied by the co sign of 30. Let's remove this right now and we have To deduct the frictional force from this and that's equal to the mass which is 70 multiplied by the exploration that we need to get. So the frictional force is as we said before, the frictional force, it's just the coefficient of kinetic friction multiplied by the normal force. So here we have opened. five, Multiplied by 487 86.7. So that's equal to 200 43.4 Newton's. So now substituting this. Here we get that the acceleration in the X direction is equal to 400 co sign 30 minus the frictional force, which is 243.4 divided by 70. And we get that. The exploration is 1.47 or approximately 1.5 meters per second squared. So that's the final answer here, and that's the exploration of the box.

Okay, so this poems fairly lengthy, but it's not too bad. So part A ah says to find the excellent components of f Blind, but says the force affluent is £20.25 degrees about the horizontal. So I guess we can call it f born instead of you. So our vector F one with just equal are magnitude of the force of 20. Uh, times CO sign the angle with the horizontal the and then 20 times sign of the angle with the horizontal. Okay, said the angle with the horizontal waas 25 degrees. So I just plug this into our calculator. So are our force. From left to right is 18 uh, point 13 Air force pulling upward. It's gonna be 8.45 those of these Aaron pounds, Okay. And then Part B, once the ex ally components of forced to so forced to, is gonna be a counter to the motion of the block. So it's only gonna be going in the leftward directions. There's F one. There's f two, so it will not be pointing up or down. So it is only gonna be pointing to the left. So let's do negative. And it was £5 and zero going up and down. Okay. Part C says these answers to express the vector sum F one plus f to in terms of the components. So we have to do is add f one F two. So it's subtracting five from 18.13 and so f one plus F to a resulting force is gonna be equal to 13.13 and the original. Why'd I have just 8.45? Okay, Part D says, give the Manchin direction of each of the other forces acting on the box. Okay. So Okay, so I think it just means, like, for F one. So the magnitude of F one would just be the square root of the sum of the square of its components. So we have 18.13 square plus 8.45 square eso. That's just calculated work there. So 18.1 dream in 8.45 that total force of, um give that it's a skid. The magnitude and direction of each of the other forces acting on the box. Okay. May positive. Okay, so I think that this is asking for undo what we did, because that's just gonna come out to be our original 20. So it's not talking about F one and F two. It's talking about the force due to gravity and the normal force. Okay, so the four state of gravity, Um oh, we'll call it G. We know that the body, the box according to the problem, weighs £40. So it's just gonna be the magnitude is gonna be £40 and in the direction because we're the moment it of gene in the direction for gene will be negative. 270 degrees are sorry. 270 degrees. This pointing downward. Let me have a normal force. Which is it gonna be? Normally it would be equal to gravity, but because we're pulling up with a force of 8.45 it's just gonna be the difference. So that's gonna be 31 0.55 pounds and then the angle of the normal force we'll be straight upward said 90 degrees. Okay. And then so the last part find the 19 direction of the net force acting on the box eso the net force in the upward direction should all cancel out, cause gravity going down is 40 and normal. Going up in the force from four. Swan. Going up would be 40. Um, so there is no upward force and somebody force and net once magnitude and direction. Okay, so it's gonna be 13.13 in the X direction and zero in the Y direction. So the magnitude of the net Force, I was just gonna be £13.13 and the angle will be zero degrees. Okay, Thank you very much.

Hello. In this problem we have the box whose mass is 60 kg. Put it on the surface and is moving with constant velocity on this surface. And as we said, the mass of this box is 60 kilograms. And the force or the horizontal force to be specific used to do this motion is equal to 114. Newton's. The mission here is to calculate the coefficient of friction. And of course in that case the coefficient of kinetic friction because the box is already moving. So we know that the coefficient of friction is the ratio between the force of friction divided by the normal force. So from here we know that the exploration of the object in the X direction is equal to zero because the object is moving with constant velocity. So from here we know that the total forces or the sum of forces and the x direction is equal to zero and that's equal to the force of pool or the force applied here minus the frictional force. The frictional force taking a negative sign being opposite to the direction of the force applied. So from here That's equal to zero From where the force of friction is equal to the force Which is equal to 140. Noten's that's the magnitude of the frictional force. Now let's consider that equilibrium in the y direction as well. Here there is no motion in the Y direction at all. So this means that the total forces acting in the Y direction is also zero. And here we have the force of weight pointing downward as usual, in the direction of Gregory. And here upwards we have the normal force that the surface acts on the body as a reaction. So from here we have again the some of the summer forces in the Y direction is equal to zero, which is equal to let's take the normal force to be positive and the way to be negative pointing in the opposite direction. So from here we have the death force, or the normal force is equal to the force of weight. And we can calculate the force of weight by multiplying the mass of the object times the gravitational acceleration, Which is 60 kg, multiplied by 9.81. Mhm. And we have the normal force of 588.6. Newton's. The final step is to substitute here in that equation to get that coefficient of kinetic friction. So from here The coefficient of kinetic friction is equal to the frictional force which is 140 Divided by the normal force which is 588.6. And this gives us the answer. And we will represent it up to three significant figures. And it's open to 38. And that's the answer for this question. And that's the coefficient of kinetic friction

My discussion. We're looking at a similar example to example seven. I told 5.7, uh, in which we have a We have two blocks off mess resting on the table and you have in external force acting from the left and pushing the blocks together. Now hear this. No idea. The blocks in contact with each other's We have a contact Force P Now the only difference between this and 5.7, it's that there is friction that we have to account for and not it. There's two different kind off, two different coefficients or friction depending on the block. So M one has coefficient off kinetic friction off you. One rest for rental is mutual. I'm gonna draw first of all the individual diagrams for each other blocks. So they started. Look, one m one. So what experiences is the external force? Of course, coming from the left, you know, so extra inserts the fictional force because of you Assume that it's gonna move towards the right. Right? So it s fictional force. Opposing is motion. If and you have gravity every every normal force. Finally, we have tea Contact force, actually contact force from M two will be sighing for someone from the right hand side toasty left. So a more accurate representation like this Course it dissecting from the itch nights from the right, It on the other end for him too. The smaller block again experiences fiction this well And from the left inside you experienced e pushing force from RM one. Basically it's the contact for speak and also the mess. I mean the weight and the normal force the usual forces. So this for him too. Not a funny net force on the system off two blocks, right? So you're considering the entire block. Here s ah, a single system. So what are the Net forces acting on? The system will be for sex on a force f as well Esti fictional forces you want from me when I'm you. Now we know that the fictional forces sexually related to the coefficient like by the normal force. So the total fiction we have a new one and the normal force from the M one. It's just equal study. He wait right to bed. And so all the forces in the Y direction this m one time street with us The fictional force for the second block and to times two. Therefore the net force considering the right side STD positive X direction then the net force would just be f minus away. The daughter frictional force. Next, you want to find what is the net force for m one and forceful and to respectively So, using the free body diagrams we can very easily find mighty Why direction their balance don't 19 or more and the rich So the only thing s t ah horizontal forces along the X direction. So again, positive direction is twisty rights. So the Net Force 40 and one b f minus p my nasty fictional false you one m one g for the m two the case off into the net Force B P minus a fictional force which is mewtwo m two g. So what we do, right? All the u turns second law for these equations for M one and m two. And it was a Loy's. Basically, based on the acceleration and the acceleration to get from the Net force equation, we just divide by the mess. So, for your 1st 1 first equation for m one e nets acceleration. Just if 0.1. It's Peter of I m one minus the one. You you are heading for the 2nd 1 creation. The net acceleration for him too. We just invite this by and to No, because the both off the blocks are moving as a single object by them. We gotta they must share the same acceleration. So actually, both off this have the same acceleration, which you can't just called it a Now we got to find what it's a You think these two equations So, uh, it's quite simple. First, you on a substitute for a GOP and the younger one p using the second equation a wish your we're gonna do is gonna find in terms off p. He scores to eight times two plus you two and two. So substituting it to this question here you have If the m one minus p that p iss, it's toe a to plus you two and two g voted by and one when I see one g close to a. Not to simplify this expression would be again modifiable sites by m one. And I'm gonna bring the term with a two d my hands site so you can get a It's just ef minus you one time you to m two G minus one and one g Put it by in one plus and no, we can find Ah, what iss de apply force. So you want to find the contact force, Pete, And we have this p or you. So what we're gonna do is just suffered you in our A do this equation nights or desist e expression for people.


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