Why Does My Calculation of Gravitational Force Keep Showing as Incorrect?

AI Thread Summary
The discussion revolves around calculating the net gravitational force on a mass at the origin due to two other identical masses positioned along the x-axis. The user initially calculated the forces from each mass but incorrectly applied the formula by subtracting instead of adding the magnitudes, leading to an incorrect final answer. It was clarified that the net force should consider the direction of the forces, resulting in a negative value. Additionally, the importance of significant figures was highlighted, as the final answer should be expressed to three significant figures. The correct net gravitational force was ultimately determined to be -0.00151 N.
saugei
Messages
4
Reaction score
0
Three identical masses of 650 kg each are placed on the x-axis. One mass is at -13.0 cm, one is at the origin, and one is at 42.0 cm. What is the magnitude of the net gravitational force on the mass at the origin, due to the other two masses?

I solved this problem in the homeowrk section of the textbook and got the answer right. This is the same problem, but with different numbers. Somehow when I enter the answer in the computer it keeps on saying that I'm wrong. Please look at my work and see what went wrong. Heres my work:

I used the formula: F_g = (Gm_1m_2)/R^2 where G= 6.67 x 10^-11

The gravitational force from the mass at -.13 m on the mass at the origin is calculated to be: -.001668

The gravitational force from the mass at .42 m on the mass at the origin is calculated to be: .00016

Net gravitational force is: |F_g1 - F_g2| = .001508 N

This answer was wrong. Thanks to anyone that is gracious enough to spend time on this problem.
 
Physics news on Phys.org
well, you're adding to get your answer, while your equation is showing the difference of the two forces.

i would think that the sum would be correct, though.
 
The sum is not correct, I tried it. The two forces are pulling in opposite directions, so I think subtracting is much more appropriate to get the "net force" on the mass at the origin.
 
Does the computer care about sig figs? Because you're only given the mass to two sig figs, whereas you're representing your answer to 4.

--J
 
Teclo and Justin both of you were right! Thanks. The net force would be to the left hence negative and its to 3 sig figs. So the answer was -.00151 compared to my original: +0.001508. Thanks
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top