What is the Correct Calculation for Force in a Three-Mass System?

[MitsuShai]

I can't find what I'm doing wrong in this problem:


Equation: f= (Gm1m)/r^2

F1= (G*(8.5))/(.2)^2 = 1.417 x 10^-8
F2= (G*(12))/.5^2 = 3.2016x10^-9

Fnet= -1.417 x 10^-8 + 3.2016x10^-9 = -1.1x10^-8m

a= 1.1x10^-8, and this is wrong and I can't figure out where I messed up at.

 

[JaredJames]

Well firstly, on F2, it should be 0.3^2 not 0.5^2 as the point is released 20cm from the 8.50kg mass which makes it 30cm from the 12.0kg mass.

Secondly, force is measured in Newtons (kg*m / s^2). Not metres (m) as you have it. You don't know the mass of particle m so you can't use it. You need the following.

g = GM / r^2 = (m^3 / kg*s^2)(kg) / m^2 = m/s^2

G = 6.67x10^-11m^3 / kg*s^2
M1 = 8.5kg
M2 = 12.0kg
r1 = 0.2
r2 = 0.3

g1 = GM1/r1^2
g2 = GM2/r2^2

g = g1 - g2

Jared

 

[MitsuShai]

Well firstly, on F2, it should be 0.3^2 not 0.5^2 as the point is released 20cm from the 8.50kg mass which makes it 30cm from the 12.0kg mass.

Secondly, force is measured in Newtons (kg*m / s^2). Not metres (m) as you have it. You don't know the mass of particle m so you can't use it. You need the following.

g = GM / r^2 = (m^3 / kg*s^2)(kg) / m^2 = m/s^2

G = 6.67x10^-11m^3 / kg*s^2
M1 = 8.5kg
M2 = 12.0kg
r1 = 0.2
r2 = 0.3

g1 = GM1/r1^2
g2 = GM2/r2^2

g = g1 - g2 (in the direction of the larger mass

Jared

oh I didn't think of that, thanks Jared