Equilibrium Of Concurrent Forces

Question

Three coplanar forces act simultaneously at a single point and hold a small ring in equilibrium. If two of the forces are 6 N pointing east and 8 N pointing north, what must the third force be?

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For a particle in equilibrium under concurrent forces, the vector sum of all forces must be zero, so the third force must exactly cancel the resultant of the other two. The 6 N east and 8 N north forces are perpendicular, giving a resultant of magnitude $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10	ext{ N}$ directed into the north-east quadrant. To balance this, the third force must have the same magnitude, 10 N, but point in the opposite direction, namely toward the south-west. The 14 N north-east option simply adds the magnitudes arithmetically, ignoring that the forces are perpendicular and that the balancing force opposes the resultant. The 10 N north-east choice has the right size but the wrong direction, so it would double the imbalance instead of cancelling it. The 2 N east option mistakenly subtracts the magnitudes. This is the NCERT triangle-law treatment of three-force equilibrium. As a check, the three vectors form a closed triangle, confirming zero net force.

Equilibrium Of Concurrent Forces

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