Answered: At point A (4.0) a mass of 3 kg is…

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Question

At point A (4.0) a mass of 3 kg is placed and at point B (8.0) another mass of 5 kg is placed. Calculate the resultant force acting on a third mass of 7 kg when it is placed at the origin of coordinates and when it is located at point C (1,3).

Expert Solution

Step 1

Given,

masses are,

$m_{1} = 3 (k g) m_{2} = 5 (k g)$

where mass m₁ is placed at point A = A(4,0). and mass m₂ is placed at point B = B(8,0).

A third mass m₃ is placed at origin "O". where m₃ = 7(kg).

The distance between m₁ and m₃ will be, ${\vec{r}}_{31} = (0 - 4) \hat{x} = - 4 \hat{x}$

And the distance between m₂ and m₃ will be, ${\vec{r}}_{32} = (0 - 8) \hat{x} = - 8 \hat{x}$

So the force on m₃ due to m₁ will be,

$\begin{array}{rcl} {\vec{F}}_{31} & = & - \frac{G m_{1} m_{3}}{{|{\vec{r}}_{31}|}^{2}} {\hat{r}}_{31} \\ = & - \frac{6.67408 \times 10^{- 11} (N . m^{2} / k g^{2}) \times 3 (k g) \times 7 (k g)}{16} (- \hat{x}) \\ = & 8.75 \times 10^{- 11} (N) \hat{x} \end{array}$

the force on m₃ due to m₂ will be,

$\begin{array}{rcl} {\vec{F}}_{32} & = & - \frac{G m_{1} m_{3}}{{|{\vec{r}}_{32}|}^{2}} {\hat{r}}_{32} \\ = & - \frac{6.67408 \times 10^{- 11} (N . m^{2} / k g^{2}) \times 5 (k g) \times 7 (k g)}{64} (- \hat{x}) \\ = & 3.61 \times 10^{- 11} (N) \hat{x} \end{array}$

So, the net force on m₃ will be,

$\begin{array}{rcl} {\vec{F}}_{3} & = & {\vec{F}}_{31} + {\vec{F}}_{32} \\ = & [8.75 \times 10^{- 11} (N) + 3.61 \times 10^{- 11} (N)] \hat{x} \\ = & 12.36 \times 10^{- 11} (N) \hat{x} \end{array}$

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