Question

three spheres of mass m1=1.11x10

Name 17' (4 pts) Three spheres of mass mi=1.11x10, kg, m2-2.22H++I -Ir x 10 kg, and m3-3.33 x 10'kg, are in deep space. TheyLt are located relative to a set of x-y coordinate axes as shown. The units for the grids on the x and y axes are kilameters (a) What is the magnitude and the direction of the net gravitational force on Sphere #37 (c) what is thework done when that same sphere is then removed from the system? ? ? A) 9.04 N, + 13.2 degrees B) 81.7 N,+76.8 degrees 3.33klu 81.7 N, +13.2 degrees 9.04 N, +76.8 degrees E) 10.9 N, +90 degrees Choices for Part (b) A) 14900 J B) +14900J -109600 J 94700 J 94700 17a.

Answer 1

Angle between line connecting 3 to 1 and the horizontal is $\theta_1 = tan^{-1}\frac{5}{6} = 39.805\degree$

and the angle between line connecting 3 to 2 and the horizontal is: $\theta_2 = tan^{-1}\frac{6}{5} = 50.194\degree$

so, the net force along x axis on 3 is:

$F_{x} = G[\frac{(3.33\times 10^9)(2.22\times 10^9)cos50.194\degree}{6000^2+5000^2} - \frac{(3.33\times 10^9)(1.11\times 10^9)cos39.805\degree}{5000^2+6000^2}]$

=> F_x = 2.0696 N

and similarly, the net force along y axis is:

$F_{y} = G[\frac{(3.33\times 10^9)(2.22\times 10^9)sin50.194\degree}{6000^2+5000^2} + \frac{(3.33\times 10^9)(1.11\times 10^9)sin39.805\degree}{5000^2+6000^2}]$

=> F_y = 8.797 N

the magnitude of the net force is then:

$F = \sqrt{F_x^2+F_y^2} = 9.037 N$

and the direction counter-clockwise from positive x axis is:

$\phi = tan^{-1}\frac{F_y}{F_x} =76.761\degree$

B] Work done in removing the sphere from the system = work done in adding the same sphere to the two-body system.

So, the work done is:

$U_i = G[\frac{1.11(3.33)\times (10^9)^2}{\sqrt{5000^2+6000^2}} + \frac{2.22(3.33)\times (10^9)^2}{\sqrt{5000^2+6000^2}}]=94699.87J$ [option D].