Forces Assignment Questions

[whyisad]

Homework Statement

There's an equilateral triangle with three objects on every corner. Corners named A, B, and C.
The mass of each:
A - 5.5 kg
B - 7.5 kg
C- 10.0 kg

The sides of the triangles are also 5.0 cm each.

There are no other forces acting on the corners. How do I calculate the net force experienced by A due to the other two objects, on corners B and C?

Homework Equations

Fnet = MA

The Attempt at a Solution

I couldn't really figure out the question... the only things I did were:
Used Fnet=MA to convert the mass of each corner into Newtons
I also realized it was an equilateral triangle, so each corner would be 60 degrees.. and therefore, we could find the exact direction that each corner is pulling towards... not sure if its relevant. They didn't give a mass for the entire triangle either, so I don't think I can find the total mass including the objects at the corners.

 

[enkerecz]

Are we talking about the force of gravity between the objects?
Or are the objects connected so there will be some tension force on one object due to the other two?

 

[whyisad]

The forces of gravity between the objects, likely.

 

[enkerecz]

Draw the triangle first and look at it.
You should notice that point masses at either end of the base of the triangle have a force acting at 60 degrees from the top mass, and a parallel force coming from the opposing end of the base. Calculate the force of gravity between each mass, paying attention to the angle it's acting at, and split it into its components. (x)i +(y)j=Fg in vector form. Pay attention the direction and just add up the components that way. Magnitude of the vector is sqrt[(x)i^2+(y)j^2]

 

[rwisz]

In order to calculate the net force experienced by corner A, we need to first find the individual forces acting on corner A from corners B and C. To do this, we can use the formula for gravitational force:

F = G * (m1 * m2) / r^2

Where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

Since we know the masses of the objects at corners B and C, we can calculate the force of gravity acting on corner A from each of them. Since the triangle is equilateral, the distance between corner A and corners B and C is the same (5.0 cm). So the formula becomes:

F = G * (m1 * m2) / (5.0 cm)^2

Now we can plug in the values:

For corner B:
F = (6.67 x 10^-11 Nm^2/kg^2) * (5.5 kg * 7.5 kg) / (5.0 cm)^2
F = 2.49 x 10^-8 N

For corner C:
F = (6.67 x 10^-11 Nm^2/kg^2) * (5.5 kg * 10.0 kg) / (5.0 cm)^2
F = 3.57 x 10^-8 N

Now we can use vector addition to find the net force experienced by corner A. Since the triangle is equilateral, the forces from corners B and C will be acting at 60 degrees from each other, in opposite directions. This means we can use the Pythagorean theorem to find the magnitude of the net force:

Fnet = √(F1^2 + F2^2 + 2F1F2cosθ)

Where F1 and F2 are the individual forces and θ is the angle between them. In this case, θ = 60 degrees.

So the net force experienced by corner A is:

Fnet = √((2.49 x 10^-8 N)^2 + (3.57 x 10^-8 N)^2 + 2(2.49 x 10^-8 N)(3.57 x 10^-8 N)cos60)
Fnet = √(6.22 x 10^-16 N^2 + 1.