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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

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Chapter 13

Practice Test 3

Section 1

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Hi in the given problem land off the mosque less bought. He'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'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.

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