Magnitude of net Gravitational Force

  • #1

Homework Statement


a) What is the magnitude of the net gravitational force on the m2=10kg mass? Assume m1=20kg and m3=15kg.
b) What is the direction of the net gravitational force on the m2=10kg mass? Assume m1=20kg and m3=15kg.

Homework Equations


F = G*m2*m3/R^2

The Attempt at a Solution


M2 x vector force
sin(26.565)*G*m2*m3/(0.223)^2
M2 y vector force
cos(26.565)*G*m2*m3/(0.223)^2

Total force
x vector
sin(26.565)*G*m2*m3/(0.223)^2 = 9.03796*10^-8
y vector
G*m2*m1/0.2^2 + cos(26.565)*G*m2*m3/(0.223)^2
= 5.0058*10^-7

magnitude
sqrt((9.03796*10^-8)^2+(5.0058*10^-7)^2) =
5.0867

angle
Don't know
 

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Answers and Replies

  • #2
collinsmark
Homework Helper
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Homework Statement


a) What is the magnitude of the net gravitational force on the m2=10kg mass? Assume m1=20kg and m3=15kg.
b) What is the direction of the net gravitational force on the m2=10kg mass? Assume m1=20kg and m3=15kg.

Homework Equations


F = G*m2*m3/R^2

The Attempt at a Solution


M2 x vector force
sin(26.565)*G*m2*m3/(0.223)^2
M2 y vector force
cos(26.565)*G*m2*m3/(0.223)^2

Total force
x vector
sin(26.565)*G*m2*m3/(0.223)^2 = 9.03796*10^-8
y vector
G*m2*m1/0.2^2 + cos(26.565)*G*m2*m3/(0.223)^2
= 5.0058*10^-7

magnitude
sqrt((9.03796*10^-8)^2+(5.0058*10^-7)^2) =
5.0867
You seem to be on the right track, but before we go any further, in what units is that answer? Your instructor may very well ding your answer for not specifying units. A simple number means little to nothing without units.

Beyond that, your answer may have some significant rounding errors. Try to work the problem symbolically before plugging the numbers in. This not only makes the problem simpler and easier to work with, but it also reduces rounding errors by reducing the number of numerical calculations.

Some useful identities are:

[itex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/itex]

[itex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/itex]

[itex] \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} [/itex]

angle
Don't know

For the angle, you'll usually use an inverse trigonometric function such as [itex] \mathrm{arctan} \left( \frac{y}{x} \right) [/itex]. But it's up to you to draw out the triangle and specify the correct angle in the triangle for which you are looking. You'll need to specify what the angle is relative to ("relative to the positive x-axis," for example).

[Edit: It's also possible to specify directions with unit vectors. I'm not sure what your instructor wants, angles or unit vectors. I'm guessing an angle.]
 
Last edited:

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