Law of gravitation and net gravitational force

Click For Summary

Homework Help Overview

The discussion revolves around the application of the law of gravitation to calculate the net gravitational force acting on a central sphere due to four surrounding spheres. The problem involves understanding vector addition of forces and the conversion to unit-vector notation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the positions of the masses and the calculation of forces. There are attempts to clarify the vector nature of gravitational forces and how to combine them correctly. Questions arise about the correct approach to find the components of the net force and the implications of angles in the calculations.

Discussion Status

Participants are actively engaging with the problem, offering insights and corrections regarding the vector addition of forces. Some guidance has been provided on how to find the components of the forces, but there is still confusion about the calculations and the correct application of trigonometric functions.

Contextual Notes

There is mention of a missing figure that is crucial for visualizing the problem setup. Additionally, participants are navigating issues related to the use of radians versus degrees in their calculations.

jperk980
Messages
47
Reaction score
0
[SOLVED] Law of gravitation

In Figure 13-34, a square of edge length 15.0 cm is formed by four spheres of masses m1 = 5.00 g, m2 = 4.00 g, m3 = 1.50 g, and m4 = 5.00 g. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.10 g?

I know that since the mass of 1 and 4 are equal they cancel out and i know that the radius is .075*Square root(2). which gives me .106m. I use the equation Gm2m5/r25^2+Gm3m5/r35^2 and i get the answer 6.85e-14 which is not the right answer. I know that the masses are .004, .0015, and .0021 since they are in grams. Can some one help me
 
Physics news on Phys.org
(1) You didn't provide the figure, but I assume that m1 and m4 are at adjacent corners not opposite corners. So they don't cancel out.
(2) The forces are vectors and must be added as such. Direction counts!
 
sorry Doc Al M1 and M$ are on opposite corners here is the picture given
 

Attachments

  • 13-34.gif
    13-34.gif
    1.6 KB · Views: 507
OK. Realize that the forces due to m2 and m3 point in opposite directions.
 
that should mean i should subtract their forces but i tried that and that did not work i got a value of 3.11e-14 which is also incorrect
 
I get the same answer. But that's the magnitude of the force. They asked for the force in unit vector notation, which means they want x and y components.
 
But how would i do that I know how to go from unit vector notation but i do not understand how to do the opposite
 
Just find the x and y components of the net force. What direction is that force? What angle does it make with the x-axis?
 
It is moving towards mass 2 and the angle should be 45
 
  • #10
jperk980 said:
It is moving towards mass 2 and the angle should be 45
Right. So what are the x and y components of a vector at an angle of 45 degrees to the horizontal?
 
  • #11
so i set up the sin 45=o/h and solve so i get hsin 45=0 and i get 2.64e-14 but that is not the correct answer. i do the same thing with cos and get 1.63e-14. where should i go from here
 
  • #12
If a force F makes an angle [itex]\theta[/itex] with the x-axis, its components are:

[tex]F_x = F \cos \theta[/tex]

[tex]F_y = F \sin \theta[/tex]
 
  • #13
Isn't that what i did, i just wrote it in different terms
 
  • #14
jperk980 said:
Isn't that what i did, i just wrote it in different terms
How did you get different answers? ([itex]\sin 45 = \cos 45[/itex].)
 
  • #15
ooo i am in radian mode
 
  • #16
Thank you Doc Al for your help
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K