Respuesta :
Answer:
[tex]F_2=2.43\times10^{-19}\ N[/tex]
[tex] \theta=21.496^{\circ}[/tex] in the anticlockwise direction from the positive x- axis.
Explanation:
Given are the masses of various spheres with their names in subscript:
- [tex]m_A=200\ kg[/tex]
- [tex]m_B=250\ kg[/tex]
- [tex]m_C=1700\ kg[/tex]
- [tex]m_D=100\ kg[/tex]
Now their respective coordinate positions (in cm) are as given below:
- [tex]P_A=(0,50)[/tex]
- [tex]P_B=(0,0)[/tex]
- [tex]P_C=(-80,0)[/tex]
- [tex]P_D=(40,0)[/tex]
Now force on B due to A:
[tex]F_{BA}=G\times \frac{200\times 250}{50^2}[/tex]
[tex]F_{BA}=20G\ N[/tex]
Now force on B due to C:
[tex]F_{BC}=G\times \frac{1700\times 250}{80^2}[/tex]
[tex]F_{BC}=66.406G\ N[/tex]
Now force on B due to D:
[tex]F_{BD}=G\times \frac{100\times 250}{40^2}[/tex]
[tex]F_{BD}=15.625G\ N[/tex]
We observe that the forces due to masses C&D act opposite in direction.
So, the net force in the x-direction:
[tex]F_x=F_{BC}-F_{BD}[/tex]
[tex]F_x=66.406G-15.625G[/tex]
[tex]F_x=50.781G\ N[/tex] in the positive x-direction
We have only one force in y-direction due to mass A.
So,
[tex]F_y=20G\ N[/tex] in the positive y-direction.
Now the net force:
[tex]F_2=\sqrt{F_y^2+F_x^2}[/tex]
[tex]F_2=\sqrt{(20G)^2+(50.781G)^2}[/tex]
[tex]F_2=54.5776G^2\ N[/tex]
[tex]F_2=2.43\times10^{-19}\ N[/tex]
Now the direction of this force with respect to x-axis:
[tex]tan\ \theta=\frac{F_y}{F_x}[/tex]
[tex]tan\ \theta=\frac{20G}{50.781G}[/tex]
[tex] \theta=21.496^{\circ}[/tex] in the anticlockwise direction from the positive x- axis.
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
[tex]F_{A}=\frac{Gm_{A}m_{B}}{0.5^{2}}[/tex]
[tex]F_{A}=\frac{6.67\times10^{-11}\times200\times250}{0.5^{2}}[/tex]
FA = 1.334 x 10^-5 N
Force on B due to C
[tex]F_{C}=\frac{Gm_{C}m_{B}}{0.8^{2}}[/tex]
[tex]F_{C}=\frac{6.67\times10^{-11}\times1700\times250}{0.8^{2}}[/tex]
FC = 4.43 x 10^-5 N
Force on B due to D
[tex]F_{D}=\frac{Gm_{D}m_{B}}{0.5^{2}}[/tex]
[tex]F_{D}=\frac{6.67\times10^{-11}\times100\times250}{0.4^{2}}[/tex]
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
[tex]F=\sqrt{F'^{2}+F_{A}^{2}}[/tex]
[tex]F=\sqrt{1.334^{2}+3.39^{2}}\times 10^{-5}[/tex]
F = 3.64 x 10^-5 N
Thus, the net force on B due to other is 3.64 x 10^-5 N.
