Answer
(a) $F_A=5.67\times10^{-5}N$, directed to the right.
(b) $F_B=3.49\times10^{-5}N$, directed to the right.
(c) $F_C=9.16\times10^{-5}N$, directed to the left.
Work Step by Step
With $m_A=363kg, m_B=517kg$ and $r_{AB}=0.5m$
$$F_{AB}=F_{BA}=G\frac{m_Am_B}{r^2_{AB}}=5.01\times10^{-5}N$$
With $m_A=363kg, m_C=154kg$ and $r_{AC}=0.75m$
$$F_{AC}=F_{CA}=G\frac{m_Am_C}{r^2_{AC}}=6.63\times10^{-6}N$$
With $m_B=517kg, m_C=154kg$ and $r_{BC}=0.25m$
$$F_{BC}=F_{CB}=G\frac{m_Bm_C}{r^2_{BC}}=8.5\times10^{-5}N$$
(a) Both particle B and C attracts particle A to the right with forces $F_{BA}$ and $F_{CA}$ respectively. Therefore,
$$F_A=F_{BA}+F_{CA}=5.67\times10^{-5}N$$
Direction: to the right
(b) Particle A pulls particle B to the left with force $F_{AB}$, while particle C pulls particle B to the right with force $F_{CB}$. If we take rightward to be the positive direction, $$F_B=F_{CB}-F_{AB}=3.49\times10^{-5}N$$
Direction: $F_B\gt0$ so the direction is to the right.
(c) Both particle A and B attract particle C to the left with forces $F_{AC}$ and $F_{BC}$ respectively. Therefore,
$$F_C=F_{AC}+F_{BC}=9.16\times10^{-5}N$$
Direction: to the left
