Universal gravitation and inclines

[AdnamaLeigh]

There's an inclined plane with theta unknown. The frictional coefficient is 0. m1 is higher on the inclined plane than m2.

m1 = 1680kg
m2 = 152kg
Distance between the two: 11mm

At what angle of inclination will the 2nd mass begin to slide down the plane?

Normally (without 2 objects) I know that net force would have to equal 0 in order for the box to slide down. In other words, it would be Fgx - Ff = Fnet = 0.

I first started with this:

1489.6sinθ - 0 = 0 But I know the law of universal gravitation plays a part in this. I was thinking about making the 1489.6sinθ equal to the universal gravitational equation since I have all the variables.

1489.6sinθ = (Gm1m2)/(r^2)

Would this be the correct thing to do? If so, I'm confused as to why they would be equal. Thanks.

 

[hotvette]

If your physics class is calculus based, the 2nd attachment in the link below illustrates a simple, structured methodology for approaching problems like this - it even has a mass on incline example. Check it out.

https://www.physicsforums.com/showthread.php?t=93670

 

[AdnamaLeigh]

Oh no, I know how to do this type of a problem when there is a single mass. But this question is implying that m1 is exerting a gravitational force on m2 and vice versa. (How do I know this for sure? The question provides a given: G=6.67259e-11, BIG hint) That's what I'm confused about.

 

[Pyrrhus]

Could you write the problem statement?

 

[AdnamaLeigh]

Given:
g=9.8m/s^2
G= 6.67259e-11

A mass of m1=1680kg is held on a frictionless surface 11mm from a second mass of m2=152kg. The surface is slowly tilted. At what angle of inclination will the 2nd mass begin to slide down the plane?

 

[Pyrrhus]

Ok basically, you do a sum of forces like you did, the force which will counterbalance the Weight of mass 2 will be the gravitational force mass 1 exerts on mass 2.
$$\sum_{i=1}^{n} \vec{F}_{i} = \vec{Fg}_{12} + m_{2} \vec{g} = \vec{0}$$

 

[hotvette]

But this question is implying that m1 is exerting a gravitational force on m2 and vice versa

Oops, didn't see that part.

 

[AdnamaLeigh]

It's okay, the method that I posted initially was correct. I wasted time, meh.