Question

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 61.0-kg object placed midway between them. 1 .0000003383 Correct: Your answer is correct. N (b) At what position (other than an infinitely remote one) can the 61.0-kg object be placed so as to experience a net force of zero from the other two objects? 2 1.824 Incorrect: Your answer is incorrect.

Answer 1

Answer:
`F_(net)=3.383* 10^(-7)\ N`

Step-by-step explanation:

Given

Mass of first object
`m_1=225\ kg`

Mass of second object
`m_2=525\ kg`

Distance between them
`d=3.8\ m`

`m_3=61\ kg` object is placed between them

So force exerted by
`m_1` on
`m_3`

`F_(13)=(Gm_1m_3)/(1.9^2)`

`F_(13)=(6.674* 10^(-11)(225* 61))/(1.9^2)`

`F_(13)=2.5374141274×10^(−7)\ N`

Force exerted by
`m_2\ on\ m_3`

`F_(23)=(Gm_2m_3)/(1.9^2)`

`F_(23)=(6.674* 10^(-11)(525* 61))/(1.9^2)`

`F_(23)=5.920632964* 10^(-7)\ N`

So net force on
`m_3` is

`F_(net)=F_(23)-F_(13)`

`F_(net)=5.920632964* 10^(-7)-2.5374141274* 10^(-7)`

`F_(net)=3.383* 10^(-7)\ N`

i.e. net force is towards
`m_2`

(b)For net force to be zero on
`m_3`, suppose

So force exerted by
`m_1` and
`m_2` must be equal

`F_(13)=F_(23)`

`\Rightarrow (Gm_1m_3)/(x^2)=(Gm_2m_3)/((3.8-x)^2)`

`\Rightarrow (m_1)/(x^2)=(m_2)/((3.8-x)^2)`

`\Rightarrow ((3.8-x)/(x))^2=(m_2)/(m_1)`

`\Rightarrow (3.8-x)/(x)=\sqrt{(525)/(225)}`

`\Rightarrow 3.8-x=1.52752x`

`\Rightarrow 3.8=2.52x`

`\Rightarrow x=1.507\ m`