Forces of 5 N and 3 N act on a particle with an angle of 60° between them. What is the size of their resultant?
- Use R2 = A2 + B2 + 2AB cos θ.
- R2 = 52 + 32 + 2 × 5 × 3 × cos 60° = 25 + 9 + 15 = 49.
- R = √49 = 7.
Answer: 7 N
Vectors and equilibrium
NUST NET Physics · Physics · Measurement, vectors and equilibrium. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
A vector has size and direction. The fast way to handle vectors is by components: split each one into parts along i, j and k, then add or subtract the matching parts.
Two vectors can be multiplied in two ways. The scalar product A · B gives a number and is zero when the vectors are perpendicular. The vector product A × B gives a vector at right angles to both and is zero when they are parallel. Work is a scalar product; torque is a vector product.
A body is in equilibrium when the forces on it add to zero and the torques about any point add to zero.
Rules to remember
- Components: Ax = A cos θ, Ay = A sin θ (θ measured from the x-axis); A = √(Ax2 + Ay2), tan θ = Ay/Ax
- Resultant of A and B with angle θ between them: R = √(A2 + B2 + 2AB cos θ); largest A + B, smallest |A − B| (the difference of the magnitudes)
- Scalar product: A · B = AB cos θ = AxBx + AyBy + AzBz; i · i = 1, i · j = 0
- Vector product: |A × B| = AB sin θ = area of the parallelogram on A and B (the triangle is half of it)
- i × j = k, j × k = i, k × i = j; reversing the order changes the sign: A × B = −(B × A)
- Torque τ = r × F, size rF sin θ = force × perpendicular distance from the pivot (N m)
- Equilibrium: sum of forces = 0 and sum of torques = 0 (clockwise moments = anticlockwise moments)
Common mistake
Swapping the order in a vector product. j × i = −k, not +k, so in τ = r × F always write r first and keep the signs of each term.
Worked example
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