Question

Four masses 1kg, 2kg, 3kg and 4kg are placed in the coordinate frame shown below. The x and y coordinates of each mass is given in brackets. The magnitude of the gravitational force on the 3 kg mass is given by ( G = 6.67 × 10^-11 m^3/kg.s^2 )​

Answer 1

☆Concept:

》Gravitational force is kind of attractive force which acts between two masses seperated by some distance 'r'.

》As Force is a vector quantity, so the net force on any mass will be calculated by vector addition.

☆Solution:

$\boxed{\bf{ \vec{F} = \dfrac{Gm_1m_2}{r^2}}}$

•kindly see the attached image! We can mark the forces on 3kg by vectors.

•put the arrows of force according to its attractive nature.

$\rightarrow\rm \vec{F_B} = \vec{F_A}+\vec{F_C}+\vec{F_D}$

$\rightarrow\rm \vec{F_B} = \dfrac{G(1)(3)}{7^2} \hat{j}+ \dfrac{G(3)(4)}{4^2} (\hat{-j})+ \dfrac{G(3)(2)}{6^2} \hat{i}$

$\rightarrow\rm \vec{F_B} = \dfrac{G(3)}{49} \hat{j}- \dfrac{G(12)}{16} (\hat{j})+ \dfrac{G(6)}{36} \hat{i}$

$\rightarrow\rm \vec{F_B} = \dfrac{G(3)}{49} \hat{j}- \dfrac{G(3)}{4} (\hat{j})+ \dfrac{G}{6} \hat{i}$

$\rightarrow\rm \vec{F_B} =( \dfrac{G(3)}{49} - \dfrac{G(3)}{4}) \hat{j}+ \dfrac{G}{6} \hat{i}$

$\rightarrow\rm \vec{F_B} = \dfrac{135G}{196}(\hat{-j})+ \dfrac{G}{6} \hat{i}$

$\rightarrow\rm \vec{F_B} = \sqrt{(\dfrac{135G}{196})^2+(\dfrac{G}{6})^2 }$

$\rightarrow\rm \vec{F_B} = G \sqrt{0.49+0.03}$

$\rightarrow\rm \vec{F_B} = 6.67 \times 10^{-11} \times 0.7$

$\rightarrow\bf \vec{F_B} \approx 4.8 \times 10^{-11} N$

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