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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top