three-masses-are-attached-to-a-15-m-long-massless-bar-mass-1-is-2-kg-and-is-attached-to-the-far-left

Question

Three masses are attached to a 1.5 m long massless bar. Mass 1 is 2 kg and is attached to the far left side of the bar. Mass 2 is 4 kg and is attached to the far right side of the bar. Mass 3 is 4 kg and is attached to the middle of the bar. At what distance from the far left side of the bar can a string be attached to hold the bar up horizontally?
(A) 0.3 m
(B) 0.6 m
(C) 0.9 m
(D) 1.2 m

Vishal Gupta · Accepted Answer

Hi in the given problem land off the mosque less bought. He&#x27;s given us L. Is equal to 1.5 m. Yeah, we can draw this Massless bar, let it be A B. No, A particle of mass M1 has been kept here at the far left end. And this mask, everyone is given us two kg then the mask M. Two which is put at the far right. And that is given as four kg and another mass which is kept at its center. See that is given to be four kg. The land of each half portion of this bar will be led by two. Now suppose the string is attached at this point P about which it is said that the bar remains horizontal means we can assume the center of of center of mass of this bar to be situated at point B. So suppose center of mass of the bar is at upwind B, which is distant X from left end. So we can mark this distance of be from A as X. So the distance between C and B will be x minus and by two. And here that stands between B and B will be l minus X. Now, as we are considering the center of mass to be situated at point B. So taking the expression for center of mass which is given as in X axis the center of mass along X axis of three points masses M one, M two and M three actually put at X one and X two and X ray that central masses even as M one X one plus M two, X two plus entry extreme divided by and one plus M two plus M three. But as we are considering center of mass to be situated at the origin, suppose this point P is at the origin. So for X zero For M1, this is two kg Multiplied by X one. And here this excellent Corresponding to P will be negative and that will be given as -1. So here this is minus X plus M two which is four kg and X two. Yeah, This X two will be corresponding to this M2 which has given us L -X. Finally entry which is again four kg and exposition Is this c. So that will be given as negative of X -2. So this is four Kg bracket minus X -L x two, divided by 1%. Plus entry means to Plus four plus 4 kg which will be shifted towards left hand side to make it zero. So finally we have two X plus four X -L by two is equal to four l minus X. Now putting the value of L. here this is two eggs plus four X -0.75 For L x two. This is your 175 m. He&#x27;s going to four l minus X means for 1.5 minus X, expanding the bracket. This is two X plus four X plus, sorry minus 3.0 Is equal to 6.00 minus cortex. So it comes out to be 10 x equals to nine oh X equals to 0.9 m from left end. Hence, we can see here. The option. See is correct. Thank you.

Salamat Ali · Answer

center of mass for a two masses can burden his ex cm. Oh, I see him. That is a cool thio x one times in one plus six two x two times in two Derided by Moss one plus master must do so substituting the values Um, we get here. The center of mass for two masses is 0.6 to 5 to fuck Where is the center of mass for three months is seated Can be again, uh can be again found by writing the equation by writing the same thing Only aiding one more extra term which is X three oh, Don&#x27;s Mars free And here in the region who he had one more term plus mastery. So subsidiary lose We get here the news and drop Mars, which is six 0.6 room 16 meters. So this is 160.6 25 meters and the difference we get here is ah, 0.25 meter. So central mass has shifted to the left. By this amount, is there a point 0 to 5 meter to the left? So option is the right answer

Ze-Han Lee · Answer

So in this problem, we have, ah, meters thick him by a threat. So this is the middle of the middle steak. Okay, so basically, this is the 50 centimetres position. And on the 11 side, we had ah object whose mess is Ah, three kilograms. And on the only right hand side we had another object. But we do not know the mass that said his M one. Ah, and the position on this one equal 80 centimeters. So want to know the massive man and one So, uh, we want to keep this meter stick bandaged. We need to make sure that the torque on this Ah, the network on this meter stick zero so we can list equation sometimes G and times 50 centimeters. Right. So this is the torque induced by this by this by this object, this should be equal to m £1 g times the distance from the 80 centimeters to the 50 centimeters. That is 30 centimeters. Right. So this on the right inside this is that a torque induced by this object so we can see that the relation between Eminem wise I&#x27;m one equal five over three times. And because Sami, three kilograms. So there&#x27;s the five kilograms. Okay,

Ben Nicholson · Answer

problem. 11.1. We have a bar that&#x27;s 50 centimetres long and has a mass with a couple of other masses. Intuit, whose size is such that we don&#x27;t need to worry about how big they are, we can treat them as his point masses. And so we want to know if we want to balance this rod with some sort of full crew where along it we would want to put the fulcrum, which is the same was asking, Where is its center of gravity? And we want. It is a length from the left. So when the variation in the force of gravity with you know elevation could be ignored, like in most cases the center of gravity in the center of Mass were the same and the make that look like a signal, the, um, center of mass is just the weighted average of all of the positions of things in the in the system. No. So you some of the massive, each object times it&#x27;s positioned. That&#x27;s all divided by the total mass. So in this case since we&#x27;re measuring, we wanted from the left end. So let&#x27;s go ahead and just, you know, measure things from from the left end. So here it would be zero, the center of mass of the bar, which we can use as its position is in the middle of it exits uniform. And then this is 50 centimeters away. So we have zero for this one. We don&#x27;t need to really worry too much about that right now. Uh and then we have are what, 0.120 kilograms of the bar times the 25 centimeters it is away from the left end. And then our 0.11 kilograms of the red mass here times half a meter away. It iss. And then all of this will be divided by the sum of the masses. 12 kilograms. Scare of playing girl, five by kilograms, plus 0.11 kilograms. And so this works out to be 29.8 centimeters. Since this is originally given in centimeters, might might as well make sure answers and seven meters to it doesn&#x27;t really matter for this. It&#x27;s as long as you don&#x27;t get confused. With what? You know how you&#x27;re using the units? Yeah, you could. Didn&#x27;t you report this in meters? Are do this calculation and centimeters. You can convince yourself that, you know, you&#x27;ll get the same answer if all of these air in centimeters instead of meters, because you&#x27;re just multiplying by 100 or dividing by 100 good. Anyway, So, yeah, So you would want to put it, you know, somewhere about here ish? Yeah. To the right of the center of the bar.

Keshav Singh · Answer

in this problem. On the topic of equilibrium and elasticity, we have shown a non uniform bar being suspended, addressed in a horizontal position by two massless cards. The first card makes an angle of 36.9 degrees with the vertical and the second card makes the angle of 53.1 degrees with the vertical. If we are given the length of the bar L. To be 6.1 m, we want to calculate the distance X from the left end of the bar to the bars center of mass. Now the bar is in equilibrium, so the forces and the talks acting on it each some 20 So if we let tlv detention force of the left hand cord and T. R. B, detention force of the right hand cord and um the mess with the bar, we can write the equations of equilibrium as follows firstly for the vertical force components. The equilibrium equation is tl coarsened data plus T. R. Co sci fi minus the weight of the bar. Mg was equal to zero Next. For the horizontal force components we have minus tl scientific data plus T R. Signed data was signed fire rather is equal to zero. Lastly, if we to create the talks we get em G times X minus tr times al times co sign fi must equal to zero. Now the origen here was chosen to be at the left end of the bar so to allow for the calculating of the talk, the unknown quantities here at T L. T. R. X. If we can eliminate T. L. And T are we can solve X as required. Now the second equation here yields tL is equal to tr signed fi divided by scientist to if we substitute this into the first equation, we get T. R. Is equal to MG. Scientific to divided by sine phi cosign theta. Less Costa and fi sine of theta. Now this expression can then be substituted into the third equation from above. And we can solve the result for X. So X. Is equal to our into science data call sign fi divided by sci fi. Course I intend to yes. Cool sci fi. I&#x27;m the sign of theater. And so if we simplify this, this becomes L. Into sine Theta. Call sign. Fi divided by the sign of data. Let&#x27;s fight. And we get this expression by using the trigger the metric identity sign of A plus B is equal to sign a course I M B. That&#x27;s co sign a sigh and be. Now in this problem we have a special case in which peter plus five is equal to 90 degrees and therefore they sign of data. Plus five is simply equal to one. And so if we simplify the equation for X, we get X to be oh times the sine of data M. D. Co sign, or if I, which is 6.1 m times a sign of 36 0.9 degrees, I&#x27;m the co sign of 53.1 degrees, and so we find the distance X from the left end of the bar to its center of mass to be 2.2 m.

Jayashree Behera · Answer

So let&#x27;s first draw the diagram showing both the masses are the objects. So here I have it fully been here. I haven&#x27;t object with Mass and one on DDE here. I haven&#x27;t object with Mass and two and they&#x27;re connected by a string. Now let&#x27;s mark all the forces. So there should be tension on Duh he had. There should be the gravitational fools off. Empt! Oh said that will be called to enter JI. And here we must have and one g on. And what else? That&#x27;s it. My Let&#x27;s start by. So there are two ways to solve this problem. So let me just start with the basics. Basically. So Solution one. Let&#x27;s go. In fact so we&#x27;re going toe right? Newton slow for these two masses. So for the first loss, So we have the Newton&#x27;s Star Lord and over here study Newton&#x27;s second law right on over here. I keep on telling mute and start Now this is Newton second now, now the half pension acting on mosques and once we&#x27;re solving for the mark, the object m one only. So we have tension acting over here, and since the surfaces frictionless, we don&#x27;t have to go into the fictional Oh here suggested, is going to be able to m oney on and solving for the extra addiction. The hides under direction City is going to be in 18 x. And I forgot to say that there is a normal force from the surface on the object as well. On those, these two are going to be equal. So n is goingto be equal to m one g. But, uh, we don&#x27;t You don&#x27;t need to write this our song for this because you&#x27;re not going to need this. So now we got the attention. Now let&#x27;s right the same thing, Farge and do and the acceleration off these two masses are going to be seen. How do you know that? That&#x27;s because what the masses are connected by a string that is not stretchable. So the distance moved by this to object in a particular interval of time is going to be saying, which means that the velocity is going to be seen and hence the exploration is also going to be seeing. So that&#x27;s I have used the same notation, eggs and eggs for both of them. Even though over here we&#x27;re solving for the article direction. So this is actually white, so we have em to t minus t to be equal to em. Two x And the reason I&#x27;m choosing entity to be positive because the direction off the movement is downward. Therefore, the object is moving downward, and hence mpg has to be better than tea in orderto moved off. Check downward. So now I can substitute this expression off tension over here, so I&#x27;ll get em to t minus and one ex to be equal. Tow them to eggs, and that way I get X to the equal do and duty over one. Bless em to on. Let me just try down the line. Do so in one is one Katie and to his creek, 80 on dhe d can be taken to be then need a post seconds guy. So you put those values over here on calculating that part, we find a the acceleration to be 7.5 meter ballistic and squish. So the kite option is going to be option. Now there is one other matter that, um you can used to solve the problem. So that is going to be really fast. That process is really fast, but it can be a little bit tricky. You need practice to be ableto solves that to be able to use this process for complicated problems. So basically, yeah, you&#x27;re going to take this holy combination as one system. So here&#x27;s my M one and two are both a single system by boat. So basically, I&#x27;m taking in one and endure as a single system, and I&#x27;ll be solving for that system directly. Instead, off taking M one as one system into and as the other system drying the free body. Diagram off those two and solving sensor that I&#x27;ll be taking in one place and tow both these objects as one single system on. Then I&#x27;ll be solving that. So if I&#x27;m considered in these two as a single system, the tension forces become internally. So if I write the Newton&#x27;s law now, these two forces own come into account because things to our Internal Forces and Newton&#x27;s law are then rewrite Newton. Second, love your writing down the external forces so these two forces just cancel out on DDE. The next step is to see what is the external force that&#x27;s acting on the whole system. So and the external force is the one that is causing the movement. For example, you have m one g and end over here on energy as well. But the only thing that is causing the two masses to move is entity. Everyone there is not playing any role because the normal force just cancels this force and one day on their foreign money has no part in moving these two objects. So the only force in the external force that got that is causing everyone and m to remove is empty. So I can&#x27;t say that the net force acting on the whole system is empty. Oh gee, the next step is to see ah, the total how many masses are involved or how many objects are involved on dhe. Whether or not the acceleration off, both the objects are equal. So, as I said before, the acceleration is going to be cool because these two mothers are connected by by unsuitable string on dhe. Therefore, you can say that the total mass is goingto be m one less in tow on the acceleration off the whole system. So months off the system is in one place int o the net force acting on the whole system is in today and acceleration off the system is just going to be So Now we will use this, uh, three questions and we will drive down on second look so f night on the system. This equal toe marks off the system times acceleration off the system on the net. Force on the system is empty. Mars off. The system is in one pleasant do. Um, acceleration off the system is eight. This way, I get a Toby, Enter ji over and one less. And to on this expression, if you compare it, uh, solution one. You get the same expression on throwback metal as well. So again, if you substitute the values, they should come out. Toby. 7.5 meters per second squared. That matches on option. So those are the two matters for solving the problem? The first matter is going to be dealing. You see, that&#x27;s the basic method standard one. The second and 3rd 0 you need a lot of practice to be able to use this off complicated problems, but eventually you&#x27;ll catch up. So yes, the kind of bounce it is. Option. Thank you.

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A 10 kg mass is attached to the ceiling with two …

Problem 31 Easy Difficulty

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