posted by Lindsay .
Two 2.160 kg crystal balls are 0.870 m apart in the vertical direction. Calculate the magnitude of the gravitational force they exert on a 8.80 g marble located 0.260 m horizontally out from the center of a line that intersects the centers of the balls. (For your own knowledge, determine the direction of the force.)
I feel like I'm close to understanding this, but I still don't quite get it. Do I have to calculate the mag. of grav. force the crystal balls exert on each other first?
<<Do I have to calculate the mag. of grav. force the crystal balls exert on each other first? >>
No, just calculate the forces between the 0.0088 kg ball and either of the 2.160 kg balls, using the universal law of gravity. The value of R, from the small marble to either ball, is
sqrt(0.435^2 + 0.260^2)= 0.5068 m
The force vectors are the same in magnitude but act in different directions. The vertical components of the two forces (with the upper and the lower ball cancel. The gravity force vectors make an angle of arctan 0.435/0.260 = 59.1 degrees with the horizontal axis of symmetry. Only the horizontal components of the gravitaional force are additive.
Place a small wooden stick over the edge of a desk and Hit the end of the stick overhanging the table so that it flies away. How is the flight distance related to the relevant parameters? What is the condition for a maximum horizontal distance?
(6.67E-11)(2.16)(0.0088) / (0.5068)^2 = 4.94E-12 N
Is that correct?
That is the force between the mabloe and one ball. For the net force with two balls, multiply by twice the cosine of the angle I calculated
Oh you're right, I forgot that step.
So 5.07E-12 N is the net force with both balls.
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