Force of Gravity question, with Newton's Law of Gravitation

[Spartan301]

Hi, I'm new here and I hope I'm posting in the right place. I have a question regarding Newton's Law of Gravitation, and it involves the formula:

F = GMm/r^2
You know, where F = Force of Gravity, G = 6.67e-11N*m^2, then M and m are two masses, and then r is the distance between the two masses squared.

The key is telling me I have the wrong answer. The correct answer is (2.0e-9,-5.2e-10) N.

Here's the question and my work:

Three 8.00 kg spheres are located on three corners of a square. Mass A is at (0, 1.7) meters, mass B is at (1.7, 1.7) meters, and mass C is at (1.7, 0) meters. Calculate the net gravitational force on A due to the other two spheres. Give the components of the force.

Mass; 8.00 kg
Mass A Located: (0,1.7)meters
Mass B Located: (1.7,1.7)meters
Mass C Located: (1.7,0) meters

Objective: Find net gravitational force on A due to the other two spheres. Give the components of the force.

Battle Plan:
Triangulate distance between A and B
Find Force of Gravity using Newton’s Law
Triangulate distance between A and C
Find Force of Gravity using Newton’s Law

Action Report and Outcome:
Distance between A and B is 1.7 m.

F = GMm/r^2
F = [(6.67e-11 N*M^2/kg^2)(8.00 kg)(8.00 kg)]/(1.7m)^2
F = (426.88e-11 N*m^2)/(2.89m^2)
F = 147.7093426e-11 N
Sig figs: 2
1.5e-9 N

(1.48e-9 N, 1.7m)

Distance between A and C: (0,1.7),(1.7,0)
D = sqrt(x2-x1)^2+(y2-y1)^2
D = sqrt(1.7-0)^2+(0-1.7)^2
D = sqrt(1.7)^2+(-1.7)^2
D = sqrt(2.89)+(2.89)
D = sqrt(5.78)
D = 2.404163056 m
sig figs: 2
D = 2.4 m

F = GMm/r^2
F = [(6.67e-11 N*M^2/kg^2)(8.00 kg)(8.00 kg)]/(2.404163056m)^2
F = (426.88e-11 N*m^2)/5.78m^2
F = 73.85467128e-11 N
F = 7.39e-10 N

I'd really appreciate help, and let me know if I can return the favor.

-Tom

 

[collinsmark]

Hello Tom (Spartan301),

Welcome to Physics Forums!

Distance between A and B is 1.7 m.

F = GMm/r^2
F = [(6.67e-11 N*M^2/kg^2)(8.00 kg)(8.00 kg)]/(1.7m)^2
F = (426.88e-11 N*m^2)/(2.89m^2)
F = 147.7093426e-11 N
Sig figs: 2
1.5e-9 N

Okay good! you've found the magnitude of the force on mass A due to mass B.

But what about the force's direction (force on A due to B)?

Or better yet, what is the force vector (force on A due to B) in terms of its x- and y-components?

(1.48e-9 N, 1.7m)

'Not sure what you mean by that.

You'll eventually want to represent the force in terms of its x- and y- components, and both components will have units of Newtons.

Distance between A and C: (0,1.7),(1.7,0)
D = sqrt(x2-x1)^2+(y2-y1)^2
D = sqrt(1.7-0)^2+(0-1.7)^2
D = sqrt(1.7)^2+(-1.7)^2
D = sqrt(2.89)+(2.89)
D = sqrt(5.78)
D = 2.404163056 m
sig figs: 2
D = 2.4 m

F = GMm/r^2
F = [(6.67e-11 N*M^2/kg^2)(8.00 kg)(8.00 kg)]/(2.404163056m)^2
F = (426.88e-11 N*m^2)/5.78m^2
F = 73.85467128e-11 N
F = 7.39e-10 N

Okay, very nice once again! You found the magnitude of the force on mass A due to mass C.

But what about the direction (force on A due to C)? Or once again, better yet, what is the force vector in terms of its x- and y-components?

 

[Spartan301]

Hey Collins, Thank you for welcoming me, and helping.
So the force between A and B are (1.5e-9N, 0N)?
And between A and C would be ...
F = 73.85467128e-11 N
F = 7.385467128e-10 N
Which, when squared is:
F = 2.717621594e-5 N
So does that make the vector force (2.7e-5N, -2.7e-5N)?

Was that what you were going for?

-Tom

 

[Spartan301]

Isn't there some way to mix them together?
(1.5e-9N, 0N)
(2.7e-5N, -2.7e-5N)
1.5 + 2.7 = 4.2e-14
sqrt(4.2) = 2.049390153e-7 N
= 2.0e-7N
0N + -2.7e-5 = -2.7e-5
sqrt(-2.7e-5) = -1.643167673e-5/2
= -1.6e-5/2N

(2.0e-7N,-1.6e-5/2N) well the x-coordinate seems to be correct, but my exponents got jumbled in the process. y-coordinates are way off.

What's your recommendation about how to blend the two force vector quantities I came up with, if my math was correct?

-Tom

 

[collinsmark]

Hey Collins, Thank you for welcoming me, and helping.
So the force between A and B are (1.5e-9N, 0N)?

Very nice!

But don't apply the required precision until the final step. We're not finished yet, so just keep the force on A from B as (1.48 x 10^-9 N, 0 N) for now.

And between A and C would be ...
F = 73.85467128e-11 N
F = 7.385467128e-10 N
Which, when squared is:
F = 2.717621594e-5 N
So does that make the vector force (2.7e-5N, -2.7e-5N)?

No, no, something's not right there.

Take a step back. You know that the magnitude of the force vector on A from C is 7.385467128 x 10^-10 N

The magnitude of a vector is like the hypotenuse of a right triangle. Go ahead a draw the vector (an arrow) that starts at A and points toward C. Actually draw this on a piece of paper. Then label angle between this vector and the horizontal as θ. Now draw two more lines, one parallel to the x-axis and other parallel to the y-axis, such that the three lines form a right triangle, with θ as one of the angles (one of the other angles will be 90^o). You end up with three lines. The hypotenuse is the magnitude of the vector. Once side will be opposite θ and the other will be adjacent to θ.

What's the relationship between opposite/hypotenuse? And the relationship adjacent/hypotenuse?

Isn't there some way to mix them together?

Yes. Once you have both vectors in terms of their x- and y-components, you can sum the two vectors together by summing their corresponding components together.

 

[Spartan301]

<BAC = θ
Sin = opp/hyp
Cos = Adj/hyp

Sinθ = 1.7/7.39e-10 N
Sinθ = 0.230040595e10N
θ = 13.29946178e10N

and

Cosθ = 1.7/7.39e-10N
Cosθ = 0.230040595e10N
θ = 76.70053822e10N

(76.7e10N, 13.3e10N)

(76.7e10N, 13.3e10N) + (1.5e-9N, 0N) = (78.2N,13.3e10N) The correct answer is (2.0e-9,-5.2e-10) N.

I'm getting a little confused, could you give me one more hint?

 

[collinsmark]

<BAC = θ
Sin = opp/hyp
Cos = Adj/hyp

Sinθ = 1.7/7.39e-10 N
Sinθ = 0.230040595e10N
θ = 13.29946178e10N

and

Cosθ = 1.7/7.39e-10N
Cosθ = 0.230040595e10N
θ = 76.70053822e10N

(76.7e10N, 13.3e10N)

(76.7e10N, 13.3e10N) + (1.5e-9N, 0N) = (78.2N,13.3e10N) The correct answer is (2.0e-9,-5.2e-10) N.

Ummm, no. You seem to be mixing forces and lengths. You can't do that!

Let's discuss the force vector on A caused by C for now. (And ignore the force vector on A caused by B for now -- you've already got that one.)

In the "force" triangle where 7.385467128 x 10^-10 N is the hypotenuse, you don't know what the sides are yet. That's what you are tying to figure out. The sides are the x- and y-components of the force.

But you do know that this triangle is similar (i.e. same angles and relative shape) to the length triangle. The length triangle has two equal sides of 1.7 m and a hypotenuse of \sqrt{(1.7)^2 + (1.7)^2}. The key point is that the two perpendicular sides are of equal length. That means the x- and y-components of this force will be equal! (Again though, this is only for the particular force on A caused by C.)

I'm getting a little confused, could you give me one more hint?

Okay, but I'm suspecting that you haven't actually drawn out the vector on a piece of paper yet, like I requested (the vector points from A to C). It should be rather obvious for this particular problem that the angle θ is -45^o.

 

[fire-boyz705]

The force between A and C 7.39e-10N have a horizontal and a vertical component.
horizontal(or y) component = 7.39e-10 cos 45
=5.225e-10
vertical(or x) component = 7.39e-10 sin 45
=5.225e-10
as it is an isosceles right angle triangle.
so the total
horizontal(or y) component=1.5e-9 + 5.2e-10
=2.02e-9
vertical(or x) component=5.2e-10