Force Exerted by Roads of Length x & Separation x - Physics Problem

[jingu]

Homework Statement

There are two roads of length x.There separation is also x.They have mass k and p.Than what is the force exerted by one on the other.

Homework Equations

I think f=gmM/R^2 IS THE EQUATION TO BE USED.

The Attempt at a Solution

When I solved (dx)^2 was coming during integration which is not solvable for me.

 

[voko]

You have not shown us how you end up with your problem, so we cannot help.

 

[Tsunoyukami]

Please show exactly what you did so we can see where you went wrong. :)

 

[jingu]

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 with (dy)^2?...I think I am going in a wrong path.guide me.

 

[voko]

Do not use 'y' for both dm1 and dm2. Use different symbols, say 'u' and 'v'.

 

[jingu]

Do not use 'y' for both dm1 and dm2. Use different symbols, say 'u' and 'v'.

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.

 

[voko]

$$

dm_1 = \frac {m_1} {x} du

\\

dm_2 = \frac {m_2} {x} dv

$$

 

[CAF123]

You are considering the force $dF$ by the infintesimal mass element $dm_1$ on one rod (road?) on another infintesimal element $dm_2$ on the other. Then summing up all the contributions. Consider some mass element (with fractional length $du/x$) on one and another mass element (of fractional length $dv/x$) on the other.

 

[jingu]

$$

dm_1 = \frac {m_1} {x} du

\\

dm_2 = \frac {m_2} {x} dv

$$

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 studying this topic for first time...help...I am trying this question since 2 days...

 

[HallsofIvy]

If you have an integrand involving "du" and "dv", what do you think it should be integrated with respect to?

 

[voko]

The integration variables are u and v, which should be obvious from 'du' and 'dv'.

What is the topic you are studying?

 

[jingu]

The integration variables are u and v, which should be obvious from 'du' and 'dv'.

What is the topic you are studying?

Various types of forces...

 

[voko]

Various types of forces...

But your difficulty has nothing to do with the topic. You are having problems with integration, which is basic calculus.

 

[jingu]

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...

 

[voko]

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...

Grade 10? Hmm. I do not think you have studied double integrals then.

Do you know what a potential of force is?

 

[jingu]

Grade 10? Hmm. I do not think you have studied double integrals then.

Do you know what a potential of force is?

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... ...

 

[barryj]

Can you not consider the road a point mass of k and p separated by x?

 

[jingu]

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

 

[voko]

You will not get a solution in this forum, it is against the rules.

 

[Dadface]

Look again at the suggestion given by barryj in post 17. I suspect that this is the approach expected for someone in tenth grade.

 

[Tsunoyukami]

Agreed - considering them as point particles is the easiest and most reasonable approach to the problem.

However, would the distance between the two depend on the set up?

Case 1: Suppose the rods are oriented vertically and are parallel (ie. similar to considering two lines x=0 and x=10). Then the two point particles are separated by a distance d=10.

Case 2: Suppose the rods lay along a line and that their ends are separated by a distance d=10; that is imagine two pencils on a desk laying one in front of the other - then the value of r in the equation for force is not simply d.

 

[jingu]

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

 

[barryj]

(Gm1m2)\4x^2 Why 4x^2?

 

[jingu]

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

 

[voko]

How are the rods situated? Like this: | | or like this: _ _?

 

[jingu]

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

 

[voko]

Approximately correct. An exact answer requires some calculus. But it is not as difficult as I originally thought, _ _ makes things much simpler.

 

[jingu]

then how we will do using calculus..?

 

[voko]

Solve a simpler problem first. You have a rod of mass M, lying on the x-axis. Its ends are at x = 0 and x = d. Find the force from the rod on a point of mass m, also lying on the x-axis, at x = u.

 

[jingu]

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