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Step-by-Step Solution
Step 1: Understand the Problem
Three identical particles A, B, and C of mass 100 kg each are placed in a straight line such that AB = BC = 13 m. A fourth identical particle P (also of mass 100 kg) is placed at a distance 13 m from B along the perpendicular bisector of AC. We need to find the resultant gravitational force on P due to A, B, and C.
Step 2: Label the Distances
Distance between A and B = 13 m.
Distance between B and C = 13 m.
Distance from B to P = 13 m.
P is on the perpendicular bisector of AC, making AP = CP = $13\sqrt{2}\,$m (because P forms a right-angled isosceles triangle with A and C).
Step 3: Write Down the Gravitational Force Formulas
The magnitude of the gravitational force between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by:
$F = \dfrac{G\,m_1\,m_2}{r^2}$
where $G$ is the universal gravitational constant.
Step 4: Compute Individual Forces
Let each mass be $m = 100$ kg. Then:
$F_{AP} = \dfrac{G\,m^2}{(13\sqrt{2})^2}
= \dfrac{G\,m^2}{169\,\times\,2}
= \dfrac{G\,m^2}{338}.$
$F_{BP} = \dfrac{G\,m^2}{13^2}
= \dfrac{G\,m^2}{169}.$
$F_{CP} = \dfrac{G\,m^2}{(13\sqrt{2})^2}
= \dfrac{G\,m^2}{169\,\times\,2}
= \dfrac{G\,m^2}{338}.$
Step 5: Resolve Forces into Components
B is directly along the line BP, so $F_{BP}$ acts along that line. The forces $F_{AP}$ and $F_{CP}$ each make a 45° angle with BP (because P is on the perpendicular bisector of AC). Hence, the component of each of these forces along the BP direction is $F_{AP}\cos 45^\circ$ and $F_{CP}\cos 45^\circ$, respectively.
Step 6: Sum the Forces Along BP
The net force along BP is:
$F_{\text{net}} = F_{BP} + (F_{AP} \cos 45^\circ) + (F_{CP} \cos 45^\circ).$
Since $F_{AP} = F_{CP}$, we get:
$F_{\text{net}}
= \dfrac{G\,m^2}{169}
+ 2 \left(\dfrac{G\,m^2}{338} \right)\cos 45^\circ
= \dfrac{G\,m^2}{169}
\;+\; 2 \left(\dfrac{G\,m^2}{338} \right)\times \dfrac{1}{\sqrt{2}}.
Factor out $\dfrac{G\,m^2}{169}$ for simplicity:
$F_{\text{net}}
= \dfrac{G\,m^2}{169} \left(1 + \dfrac{1}{\sqrt{2}}\right).
Step 7: Substitute $m = 100$ kg and Approximate Numerically
$m^2 = (100)^2 = 10000,
\quad \dfrac{10000\,G}{169} \approx 59.17\,G,
\quad \dfrac{1}{\sqrt{2}} \approx 0.707.
Therefore,
$F_{\text{net}}
\approx 59.17\,G \times \left(1 + 0.707\right)
\approx 59.17\,G \times 1.707
\approx 101\,G.
Rounded to a convenient value, this is approximately $100\,G$.
Final Answer
The gravitational force on the fourth particle P is about $100\,G$.
