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NEET PHYSICSRotational motionMedium

Question

ABC is an equilateral triangle with O as its centre. F1\vec{F}_1, F2\vec{F}_2 and F3\vec{F}_3 represent three forces acting along the sides AB, BC and AC respectively. If the total torque about O is zero then the magnitude of F3\vec{F}_3 is:

A

F1+F2F_1+F_2

B

F1F2F_1-F_2

C

F1+F22\frac{F_1+F_2}{2}

D

2(F1+F2)2(F_1+F_2)

Step-by-Step Solution

Let rr be the perpendicular distance from the centre O of the equilateral triangle to each of its sides. The direction of torque is determined by the cross product of the position vector and the force vector. Forces F1\vec{F}_1 (acting along AB) and F2\vec{F}_2 (acting along BC) will produce torque in the same rotational sense (either both clockwise or both counter-clockwise) about the centre O. However, the force F3\vec{F}_3 is acting along AC (not CA, which would be the continuous cyclic order). Therefore, F3\vec{F}_3 will produce a torque in the opposite sense to that of F1\vec{F}_1 and F2\vec{F}_2. The total torque about O is the algebraic sum of the individual torques: τtotal=F1r+F2rF3r\tau_{\text{total}} = F_1 r + F_2 r - F_3 r Given that the total torque about O is zero: F1r+F2rF3r=0F_1 r + F_2 r - F_3 r = 0 Dividing by rr (since r0r \neq 0): F3=F1+F2F_3 = F_1 + F_2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionequilateraltrianglecentrerepresentforces

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