- #1

- 1,196

- 1

A 200 kg mass and a 500kg mass are separated by 0.4m. At what position can a 50kg mass be placed so as to experience a net force of zero (other than infinity)?

here's what I did:

I need both forces to equal each other.

[tex]F_{200}=G\frac{200m}{x^2}=G\frac{500m}{y^2}=F_{500}[/tex]

[tex]\frac{200}{x^2}=\frac{500}{y^2}[/tex]

[tex]200y^2=500x^2[/tex]

[tex]x+y=0.4[/tex]

[tex]x=0.4-y[/tex]

[tex]500(0.4-y)^2=200y^2[/tex]

[tex]500(0.16-0.8y+y^2)=200y^2[/tex]

[tex]200y^2=80-400y+500y^2[/tex]

[tex]1=\frac{0.4}{y^2}-\frac{2}{y}+2.5[/tex]

[tex]y=0.245m,x=0.155m[/tex]

did I do this correctly? was there a faster way?

here's what I did:

I need both forces to equal each other.

[tex]F_{200}=G\frac{200m}{x^2}=G\frac{500m}{y^2}=F_{500}[/tex]

[tex]\frac{200}{x^2}=\frac{500}{y^2}[/tex]

[tex]200y^2=500x^2[/tex]

[tex]x+y=0.4[/tex]

[tex]x=0.4-y[/tex]

[tex]500(0.4-y)^2=200y^2[/tex]

[tex]500(0.16-0.8y+y^2)=200y^2[/tex]

[tex]200y^2=80-400y+500y^2[/tex]

[tex]1=\frac{0.4}{y^2}-\frac{2}{y}+2.5[/tex]

[tex]y=0.245m,x=0.155m[/tex]

did I do this correctly? was there a faster way?

Last edited: