Question

In Fig 13-37 a, particle A is fixed in placed at x= -0.20 m on the x aixs and particle B with a mass 1.0 kg is fixed in place at the origin particle C are ( not shown ) can be moved along the x axis , between particle B and x= $\infty$ Figure 13-37b shows the x component $F_{netx}$ of the net gravitational force on particle B due to practices A and C, as a functional of position x particle c. The plot actually extend to the right approaching an asymptote of $-4.17\times {10}^{-10}$ N as $x\to \infty$ what are the masses of (a) particles A and (b) particles (C)

Answer 1

It is known that the value of the force, when particle C is removed (that is, as its position x goes to infinity), which is a situation in which any force caused by C vanishes. because $F\propto 1/r^2$

Thus, this situation only involves the force exerted by A on B:

$F_{net,x}=F_{AB}=\frac{G{m}_A{m}_B}{r_{AB}^2}$

It is given that this force $F_{net,x}=F_{AB}=4.17\times {10}^{-10}N$

putting values $m_B=1kg$ and $r_{AB}=0.2m$

$m_A=\frac{r_{AB}^2{F}_{AB}}{G{m}_B}=\frac{{0.2}^2\times 4.17\times {10}^{-10}}{6.67\times {10}^{-11}\times 1}=0.25kg$

(b) From Graph : that the net force on B is zero when x = 0.40 m. Thus, at

that point, the force exerted by C must have the same magnitude (but opposite direction)

as the force exerted by A (which is the one discussed in part (a)). Therefore

$\frac{G{m}_C{m}_B}{{0.4}^2}=4.17\times {10}^{-10}N$

$m_C=1.00kg$