# Find the magnitude of the acceleration of the particle.

1. Nov 28, 2007

### saturn67

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

2. Nov 28, 2007

### Galileo's Ghost

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.

Last edited: Nov 28, 2007
3. Nov 28, 2007

### saturn67

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) ???

4. Apr 8, 2008

### treverd

bump, is this equation right

a= Fg/(m2+m1)

5. Apr 8, 2008

### treverd

I figured it out

you 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.