Universal Gravity & Force Problem (should be easy)

AI Thread Summary
The discussion focuses on calculating the resultant gravitational force on a mass at the origin due to two other masses in the xy plane. The user initially calculated the forces incorrectly and was confused about vector addition, mistakenly treating the magnitudes as scalars. It was clarified that the forces are vectors and should be resolved into components before summing them. The user corrected the calculations, realizing the importance of maintaining precision to avoid round-off errors. Ultimately, the correct approach led to a successful resolution of the problem.
bmoore509
Messages
66
Reaction score
0

Homework Statement



Three 6 kg masses are located at points in
the xy plane as shown.
r1=47 cm
r2=55 cm
What is the magnitude of the resultant
force (caused by the other two masses) on
the mass at the origin? The universal gravita-
tional constant is 6.6726 × 10^−11 N · m2/kg2.
Answer in units of N.

Homework Equations


F=G(m2m1/r^2)


The Attempt at a Solution


So I got 8 x 10^-9 for one of the Forces and 1.1x10^-8 for the other. Then I took the magnitude (each squared, added together then square root) and got 1.36015x10^-8. But my online homework said it was wrong. I don't see where I went wrong.
 
Physics news on Phys.org
The resultant force is the vector sum of the two forces.

| \vec{F}_1+\vec{F}_2 |^2 \neq | \vec{F}_1|^2+ |\vec{F}_2 |^2
 
Oh, so I should have done 8 x 10^-9 + 1.1x10^-8 and then squared it?
 
bmoore509 said:
Oh, so I should have done 8 x 10^-9 + 1.1x10^-8 and then squared it?

Probably not, they're vectors so they have directions. Without being able to see the diagram, I can only suggest that you determine the components of each vector with respect to the x and y axes. Then you can compute the vector sum.
 
Well, 8x10^-9 is in the y direction and 1.1x10^-8 is in the x direction.

I'm not quite sure what you're telling me to do, though.
 
You might want to check your textbook for a discussion of vector addition, I'm not sure why you haven't covered that yet if you're already doing problems like this. In any case, check your math. I think both forces are off by a factor of 100, perhaps a result of incorrectly converting cm to m.
 
We've covered vector addition. I'm looking for the magnitude of the resultant vector. I thought that would be the square root of (F1^2+f2^2). I'm confused as to why it's not. Or what it is then.
 
47 cm = .47 m
55 cm = .55 m
G = 6.6726 × 10^−11 N · m2/kg2.

Fy = (6.6726 × 10^−11 N · m2/kg2)(6*6)/(.55^2)
=0.000000008
Fx= (6.6726 × 10^−11 N · m2/kg2.)(6*6)/.47^2)
=0.000000011

F = sqrt(0.000000011^2 + 0.000000008^2)=0.000000014
 
bmoore509 said:
We've covered vector addition. I'm looking for the magnitude of the resultant vector. I thought that would be the square root of (F1^2+f2^2). I'm confused as to why it's not. Or what it is then.

You didn't state originally that the forces were perpendicular. If they're perpendicular your calculation is correct. The factor of 100 was my mistake. The only other thing that I can suggest is that your discrepancy with what the computer is looking for is due to round-off error. Carry through more more digits and round off the final answer.
 
  • #10
I'm sorry. I should have stated that. I carried it out as far as my calculator can go. What should I do next? (The online homework never makes me round off.)
 
  • #11
I got it! You were right. So I used wolfram online to calculate it.
 
Back
Top