The resultant of these forces OP⃗,OQ⃗,OR⃗,OS⃗ and OT⃗ is approximately .......... N.
[Take √3=1.7, √2=1.4 Given î and ĵ unit vectors along x, y axis]

Solution

Physics sir · 4 steps

Break every force into x and y parts, add each column separately, and the resultant falls out.

1. Read the angles. Measure each angle from an axis. P is 30° from the y-axis and Q is 30° above the x-axis. T is 60° from the y-axis, which means 30° above the negative x-axis. R and S are each 45° below the x-axis, on opposite sides.
2. Resolve each force. The component along the axis you measured from takes cos, and the other component takes sin. Use cos 30° = √3/2 = 0.85 and cos 45° = √2/2 = 0.7. Give each component the sign of the quadrant the force points into.

   Force	Direction	x-component (N)	y-component (N)
   OP⃗, 20 N	30° from +y	20 sin 30° = 10	20 cos 30° = 17
   OQ⃗, 10 N	30° above +x	10 cos 30° = 8.5	10 sin 30° = 5
   OT⃗, 15 N	60° from +y, towards -x	-15 sin 60° = -12.75	15 cos 60° = 7.5
   OR⃗, 20 N	45° below +x	20 × 0.7 = 14	-14
   OS⃗, 15 N	45° below -x	-15 × 0.7 = -10.5	-10.5
   Sum	all five forces	9.25	5

3. Add the columns. Σ Fₓ = 10 + 8.5 - 12.75 + 14 - 10.5 = 9.25 N and Σ F_y = 17 + 5 + 7.5 - 14 - 10.5 = 5 N.
4. So. The resultant is (9.25î + 5ĵ) N, which is option (a).

The idea to keep. For many forces at odd angles, never add magnitudes. Resolve each one along x and y, attach signs by quadrant, and add the two columns on their own.

Answer from the NTA final answer key. Solution by PrepWiser.