Question

O. Two bodies of masses 8 kg are placed at the vertices A and B of an equilateral triangle ABC A third body of mass 2 kg is placed at the centroid G of the triangle. If AG = BG = CG = 1 m. where should a fourth body of mass 4 kg be placed so that the resultant force on the 2 kg body is zero? (A) at C 1 (B) at a point P on the line CG such that PG = 2. (C) at a point P on the line CG such that PG = 0.5 m v (D) at a point P on the line CG such that PG = 2m m​

Answer 1

Answer:

20 kg per minute net per har

Answer 2

Answer:

At a point P on the line CG such that PG = 1/$\sqrt{2}$ m

Explanation:

The centroid of a triangle is fashioned with the aid of using the intersection of the medians of the triangle. Medians continually lie inside a triangle. Therefore, their intersection factor i.e., the centroid of a triangle continually lies internal a triangle. The centroid is normally denoted with the aid of using 'G'.Then, we will calculate the centroid of the triangle with the aid of using taking the common of the x coordinates and the y coordinates of all of the 3 vertices. So, the centroid system may be mathematically expressed as G(x, y) = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3).

Fa=Fb= Gm1m2/r²  =G8 x 2 /1

= G(16)

Fab=$\sqrt{Fa^{2}+Fb^{2}+2FaFb cosθ }$

Fab=Fa= G(16)

For resultant force on 2kg to be zero

Fcg = -Fab

G2 x 4 / X² =G(16)

X² = 2 x 4 / 16

X² = 1/2

X=1/$\sqrt{2$

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