Find the Force of a Particle Attracted Toward Origin

[Logik]

Homework Statement

A particle is attracted toward the origin by a force proportional to the cube of it's distance from the origin. (...)

What would be this Force equal to (in xy plane)?

The Attempt at a Solution

So distance is Sqrt[x^2+y^2]... and from here I don't know what to do...

F=( ? i , ? j)

 

[Dick]

The force should also point in the direction of the origin. So consider F=k*(x*i,y*j). The magnitude of this force is directly proportional to distance from the origin. How would you change it to get the right proportionality (BTW - did you mean 'inversely' proportional to cube of the distance?).

 

[Logik]

F=k*(x^3*i,y^3*j), k<0 ?

Is this good then?

// I did not mean inversely.

 

[Dick]

F=k*(x^3*i,y^3*j), k<0 ?

Is this good then?

// I did not mean inversely.

That i) does not point towards the origin anymore and ii) is NOT proportional to distance cubed. It's proportional to sqrt(x^6+y^6). Distance cubed is sqrt(x^2+y^2)^3. Not at all the same thing. If (x,y) is already proportional to distance, why not just multiply it by distance squared?

 

[HallsofIvy]

Dick SAID you must F=k*(x*i,y*j). Here k is "proportional to the cube of it's distance from the origin". Yes, the "distance from the origin is $\sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2}$. What is the cube of that? And since the particle is "attracted", the vector must be directed toward the origin. (xi, yj) is directed away from the origin.

 

[Logik]

TO DICK.

F=(x*i, y*j)
DIV F = d/dx(x) + d/dy(y) = 2 which means it goes outward and not to the origin...

// EDIT

F=-(x^2 + y^2)^3/2 * (x*i, y*j)

 

[Dick]

TO DICK.

F=(x*i, y*j)
DIV F = d/dx(x) + d/dy(y) = 2 which means it goes outward and not to the origin...

// EDIT

F=-(x^2 + y^2)^3/2 * (x*i, y*j)

You can point it in the right direction by putting a -k in front. But we're getting there. Your latest effort points in the right direction, but (x,y) is proportional to distance and (x^2+y^2)^(3/2) is proportional to distance cubed. So the product is proportional to distance to the fourth. You just need one minor adjustment.