Question

Objects with masses of 266 kg and 647 kg
are separated by 0.372 m. A 33.3 kg mass is
placed midway between them. Find the magnitude of the net gravitational
force exerted by the two larger masses on the
33.3 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m2
/kg2. Answer in units of N.

Answer 1

Let the first mass be $m_1 = 266 kg$, the second mass be $m_2 = 647 kg$ and the third mass placed midway between them be $m_3 = 33.3 kg$. Also, let the separating distance between the first and second masses be $r_{12} = 0.372 m.$ At the midway, the forces exerted by the two larger masses on the third mass are oppositely directed. Let's choose the direction to the larger mass as the positive and write the Newton's Law of Universal Gravitation:

$$F_{net} = F_{23} - F_{13},$$ $$F_{net} = G\dfrac{m_2m_3}{(r_{23})^2} - G\dfrac{m_1m_3}{(r_{13})^2},$$

here, $F_{net}$ is the net gravitational force exerted by the two larger masses on the 33.3 kg mass, $F_{23}$ is the gravitational force exerted by the larger 647 kg mass on the 33.3 kg mass and $F_{13}$ is the gravitational force exerted by the smaller 266kg mass on the 33.3 kg mass.

Since, $r_{13} = r_{23} = \dfrac{1}{2}r_{12}$ we can write:

$$F_{net} = G\dfrac{m_2m_3}{(r_{23})^2} - G\dfrac{m_1m_3}{(r_{13})^2} = \dfrac{Gm_3}{(\dfrac{1}{2}r_{12})^2}(m_2 - m_1) = \dfrac{4Gm_3}{(r_{12})^2}(m_2 - m_1).$$

From this formula, we can find the magnitude of the net gravitational force exerted by the two larger masses on the 33.3 kg mass:

$$F_{net} = \dfrac{4 \cdot 6.672 \cdot 10^{-11} Nm^{-2} \cdot 33.3 kg}{(0.372 m)^2}(647kg - 266kg) = 2.45 \cdot 10^{-5} N.$$

Answer:

$F_{net} = 2.45 \cdot 10^{-5} N.$