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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?

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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