Question

Four uniform spheres, with masses mA = 200 kg, mB = 250 kg, mC = 1700 kg, and mD = 100 kg, have (x, y) coordinates of (0, 50 cm), (0, 0), (-80 cm, 0), and (40 cm, 0), respectively. What is the net gravitational force F2 on sphere B due to the other spheres?

Answer 1

3.64 x 10^-5 N

Step-by-step explanation:

mA = 200 kg

mB = 250 kg

mc = 1700 kg

mD = 100 kg

Force on B due to A

`F_(A)=(Gm_(A)m_(B))/(0.5^(2))`

`F_(A)=(6.67*10^(-11)*200*250)/(0.5^(2))`

FA = 1.334 x 10^-5 N

Force on B due to C

`F_(C)=(Gm_(C)m_(B))/(0.8^(2))`

`F_(C)=(6.67*10^(-11)*1700*250)/(0.8^(2))`

FC = 4.43 x 10^-5 N

Force on B due to D

`F_(D)=(Gm_(D)m_(B))/(0.5^(2))`

`F_(D)=(6.67*10^(-11)*100*250)/(0.4^(2))`

FD = 1.042 x 10^-5 N

FD and Fc are opposite to each other, so net of Fc and FD is F'.

F' = (4.43 - 1.042) x 10^-5 N towards left

F' = 3.39 x 10^-5 N

Now F' and FA are perpendicular to each other, So net force on B due to all other

`F=\sqrt{F'^(2)+F_(A)^(2)}`

`F=\sqrt{1.334^(2)+3.39^(2)}* 10^(-5)`

F = 3.64 x 10^-5 N

Thus, the net force on B due to other is 3.64 x 10^-5 N.

Answer 2

`F_2=2.43*10^(-19)\ N`

`\theta=21.496^(\circ)` in the anticlockwise direction from the positive x- axis.

Step-by-step explanation:

Given are the masses of various spheres with their names in subscript:

• 
  `m_A=200\ kg`

• 
  `m_B=250\ kg`

• 
  `m_C=1700\ kg`

• 
  `m_D=100\ kg`

Now their respective coordinate positions (in cm) are as given below:

• 
  `P_A=(0,50)`

• 
  `P_B=(0,0)`

• 
  `P_C=(-80,0)`

• 
  `P_D=(40,0)`

Now force on B due to A:

`F_(BA)=G* (200* 250)/(50^2)`

`F_(BA)=20G\ N`

Now force on B due to C:

`F_(BC)=G* (1700* 250)/(80^2)`

`F_(BC)=66.406G\ N`

Now force on B due to D:

`F_(BD)=G* (100* 250)/(40^2)`

`F_(BD)=15.625G\ N`

We observe that the forces due to masses C&D act opposite in direction.

So, the net force in the x-direction:

`F_x=F_(BC)-F_(BD)`

`F_x=66.406G-15.625G`

`F_x=50.781G\ N` in the positive x-direction

We have only one force in y-direction due to mass A.

So,

`F_y=20G\ N` in the positive y-direction.

Now the net force:

`F_2=√(F_y^2+F_x^2)`

`F_2=√((20G)^2+(50.781G)^2)`

`F_2=54.5776G^2\ N`

`F_2=2.43*10^(-19)\ N`

Now the direction of this force with respect to x-axis:

`tan\ \theta=(F_y)/(F_x)`

`tan\ \theta=(20G)/(50.781G)`

`\theta=21.496^(\circ)` in the anticlockwise direction from the positive x- axis.