Q.Three concurrent forces of magnitudes $10\ \text{N}$, $10\ \text{N}$, and $10\sqrt{2}\ \text{N}$ keep a point in equilibrium. The first two forces act at right angles to each other. Find the direction of the third force relative to the other two.

Let the two $10\ \text{N}$ forces act at right angles to each other -- say, one along the positive $x$-axis and the other along the positive $y$-axis. Their resultant has magnitude $R = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}\ \text{N}$ directed at $45^\circ$ from each of the two original forces (bisecting the right angle between them, i.e. along the direction that makes equal $45^\circ$ angles with both the $x$- and $y$-axes).\n\nFor the point to be in equilibrium under all three forces, the third force must exactly cancel this resultant -- so it must be equal in magnitude ($10\sqrt{2}\ \text{N}$, matching the given value) and exactly opposite in direction to $R$, i.e. directed at $45^\circ + 180^\circ = 225^\circ$ from the first force's direction.\n\nMeasuring the angle between this third force and each of the two original $10\ \text{N}$ forces: since the third force sits at $225^\circ$ while the …

Find the resultant of the two perpendicular forces using the Pythagorean rule, note it bisects the right angle between them, then require the third f …

• Assuming the third force is simply the sum of the other two, rather than equal and opposite to their resultant. …