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bearhunter [10]
3 years ago
6

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?
Physics
2 answers:
Anastasy [175]3 years ago
4 0

Answer:

3.64 x 10^-5 N

Explanation:

mA = 200 kg

mB = 250 kg

mc = 1700 kg

mD = 100 kg

Force on B due to A

F_{A}=\frac{Gm_{A}m_{B}}{0.5^{2}}

F_{A}=\frac{6.67\times10^{-11}\times200\times250}{0.5^{2}}

FA = 1.334 x 10^-5 N

Force on B due to C

F_{C}=\frac{Gm_{C}m_{B}}{0.8^{2}}

F_{C}=\frac{6.67\times10^{-11}\times1700\times250}{0.8^{2}}

FC = 4.43 x 10^-5 N

Force on B due to D

F_{D}=\frac{Gm_{D}m_{B}}{0.5^{2}}

F_{D}=\frac{6.67\times10^{-11}\times100\times250}{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}}\times 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.

rusak2 [61]3 years ago
3 0

Answer:

F_2=2.43\times10^{-19}\ N

\theta=21.496^{\circ} in the anticlockwise direction from the positive x- axis.

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)

<u>Now force on B due to A:</u>

F_{BA}=G\times \frac{200\times 250}{50^2}

F_{BA}=20G\ N

<u>Now force on B due to C:</u>

F_{BC}=G\times \frac{1700\times 250}{80^2}

F_{BC}=66.406G\ N

<u>Now force on B due to D:</u>

F_{BD}=G\times \frac{100\times 250}{40^2}

F_{BD}=15.625G\ N

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

<u>So, the net force in the x-direction:</u>

F_x=F_{BC}-F_{BD}

F_x=66.406G-15.625G

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

<em>We have only one force in y-direction due to mass A.</em>

So,

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

<u>Now the net force:</u>

F_2=\sqrt{F_y^2+F_x^2}

F_2=\sqrt{(20G)^2+(50.781G)^2}

F_2=54.5776G^2\ N

F_2=2.43\times10^{-19}\ N

<u>Now the direction of this force with respect to x-axis:</u>

tan\ \theta=\frac{F_y}{F_x}

tan\ \theta=\frac{20G}{50.781G}

\theta=21.496^{\circ} in the anticlockwise direction from the positive x- axis.

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