q01b
[F_{1 \rightarrow 2} = -G \frac{ m_1 m_2}{ r_{12}^2} \hat r_{12}]
For particle of equal mass [M] are fixed at the corners of a square with sides of length [a]. A fith particle has mass [m] and moves under the gravitational forces fo the other four.
Find the [x]-compnent and [y]-component of the net gravitational force on [m] due to the other foure masses when [m] is located at the center of the square [lef-handed figure]
Hint: Draw a sketch! Use superposition and draw a vector diagram consisting of four vectors, each representing the force exerted by one on the corner particles on [m]. For ease of reference, label the four [equal] corner masses [M_1, M_2, M_3, M_4]. Lable the corresponding force vectors [F_1, F_2] etc. With the fector diagram in hand, it is vastly easier to calculate the requested componets of the total force.
[\overrightarrow F = 0]
Each of the masses pull in equal and oposite directions.
Find the [x]-compnent and [y]-component of the net gravitational
force on [m] due to the other foure masses when [m] is located at
the center point of the right-hand side of the square [middle figure].
Use the same solution procedure that was recommended above for part (a)
[ F_y = 0 ]
[ F_x = - \frac{ 16 GMm}{ 5^{ \frac{ 3}{2}} a^2} ]
Check your answer to part (b) by testing at least 3 examples of limiting behavior Do you get the results you expect?
[ F_y = 0 ]
[ F_x = -2GMm \left( \frac{ x - a}{ \left( \frac{ a^2}{ 4} + \left(x - \frac{ a}{ 2}\right)^2 \right)} \frac{ x + a}{ \left( \frac{ a^2}{ 4} + ( a + \frac{ a}{2})^2 \right)^{ \frac{ 3}{2} \right) ]
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