Find the magnitude of the acceleration of the particle.

AI Thread Summary
The discussion revolves around calculating the acceleration of a particle released between two fixed masses, 8.50 kg and 13.5 kg, positioned 50 cm apart. The initial attempt incorrectly used the distance between the masses instead of the distance from the particle to each mass. The correct approach involves determining the gravitational forces acting on the particle from both masses and calculating the net force. By applying the formula for gravitational force and considering the distances from the particle to each mass, the correct acceleration can be derived. Ultimately, the net acceleration is found by subtracting the forces acting on the particle from each mass.
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An 8.50 kg point mass and a 13.5 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 19.0 cm from the 8.50 kg mass along the line connecting the two fixed masses.

Find the magnitude of the acceleration of the particle.

Fg=(Gm1m2)/r^2


G=6.67*10^-11
m1=8.50kg
m2=13.5
r=(50/2)/100=.25m

so Fg =((6.67*10^-11)*8.50*13.5)/(.25^2)
Fg=1.2246*10^-7

F=ma since particle close to 8.5kg
a= 1.2246*10^-7/8.50
=1.440*10^-8 m/s^2

why this answer wrong please help me
 
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You seem to be trying to calculate the force that exists between the 8.5 and the 13.5 kg masses (although the distance between them is 0.5 not .25m). But this question is really about mass m.


mass m is being pulled by two different forces...one from the 8.5 kg mass and the other from the 13.5 kg mass. It may help to draw a free body diagram of mass m and then write a net force equation for it. It is this net force that is responsible for the acceleration.
 
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so Fg =((6.67*10^-11)*8.50*13.5)/(.5^2) = 3.6018*^-9 N

Fnet = F2+F1= F(m2+m1)

so a= Fg/(m2+m1) ?
 
bump, is this equation right

a= Fg/(m2+m1)
 
I figured it outyou take ((G)(8.5)(m))/(.19m)^2
=1.57E-8 m

then you take ((G)(13.5)(m))/(.31m)^2
=9.37E-9 m

now just subtract the bottom one from the top to get your acceleration.
 
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