Recent content by jingu

thank you very much...Are u sure that the answer is the same...?

I solved it I came up with this answer F= (GMm ln(4/3))/d^2...please verify...

Net force on point mass m? It will depend on the positions of m2 and m bcoz gravitational force is always attractive so there will be two cases ...please tell whether u > 3d...or d<u<2d

F= (GMm)/(u^2-du) here du is the product of coordinates not differentiation. Answer is this...

then how we will do using calculus..?

they are like this _ _. My answer is correct or not...?

We will treat both of the point masses at the centre of the rod.?So distance between them is x+x/2 +x/2 = 2x...Am I wrong...Or the answer would be only Gm1m2/x^2

thank u everyone ! I think the answer is (Gm1m2)\4x^2...please verify and sorry to everyone.I was not familiar with these rules...

I don't think so...I am expecting a solution but no one is ready to give...

NO.That's why I posted it here, i hoped a solution to this problem...but u r not helping me...Give me a full solution so that i can understand it thoroughly... ...

I am not able to understand what u all are telling...?explain in detail...I am in 10 grade so Iam not familiar with calculus...Only very basic I know...

Various types of forces...

dF=(Gm1m2dvdu)/x^2*r^2 here r is the distance between them ,then r equals what...?This question would be easy if there is a point mass and a continuous body...but here both are continuous...? Then this need to integrated but with respect to what...? Please show me the calculations ...I am...

I am getting confused there are so many variables... Please give me the description of how to approach the problem.Everything else will be my work.I have to submit it tommorow.

all i did is this... Let a small element of road 1 be dy.Then dm1= (m1/x)dy.Similarly dm2=(m2/x)dy. dF=(G(dm1)(dm2))/y^2 dF=(Gm1*m2*(dy)^2)/y^2 ∫dF=∫(Gm1*m2*(dy)^2)/y^2 here the problem begins we have to integrate from x to 2x.WRT to (dy)^2...now can we integrate this... how to solve...