01 · Equilibrium
Forces are (p + 2)i + 3j N and −6i + (q − 5)j N. Find p and q for equilibrium.
Hint
Add components and set both sums to zero.
Worked solution
p − 4 = 0 and q − 2 = 0, so p = 4 and q = 2.
Understand · explore · practise
Unknown forces and vector conditions
Find unknown force components from equilibrium, parallel and perpendicular directions, and resultant magnitude while checking every algebraic root.
Before you startResultant force vectors, simultaneous equations, Pythagoras and perpendicular gradients.
01 / Translate the wording into a precise condition
Equilibrium, direction and magnitude supply different equations.
Start by simplifying the resultant.
Equilibrium requires every resultant component to be zero. A direction condition fixes the ratio of its components. A magnitude condition fixes the sum of their squares. Solve the appropriate equations, then check the original wording.
One vector, different conditionsExplore
The resultant is R = ki + 4j N. Compare its length with 5 N, its direction with 3i + 4j, and whether it is perpendicular to 2i + j. These are three different conditions, so they need not give the same k.
02 / Set each component to zero in equilibrium
Two unknown components usually need two equations.
Forces are (2p + 1)i + (q − 3)j N and (p − 7)i + (2q + 6)j N.Worked example
R = (3p − 6)i + (3q + 3)j N
Collect components first.
3p − 6 = 0 and 3q + 3 = 0
Equilibrium requires both conditions.
p = 2 and q = −1
Substitution makes both resultant components zero.
02 · Coupled unknowns
The resultant is (p + q − 7)i + (2p − q − 2)j N. Find p and q for equilibrium.
Hint
Solve the two equations simultaneously.
Worked solution
p + q = 7 and 2p − q = 2. Adding gives 3p = 9, so p = 3 and q = 4.
03 / Use a scalar multiple for parallel nonzero vectors
Parallel can mean the same or opposite direction.
Write R = λd for a nonzero direction vector d.
Compare corresponding components to find λ and any unknowns. A positive λ gives the same direction; a negative λ gives the opposite direction. Exclude λ = 0 if the question requires a direction, because the zero resultant has no direction.
03 · Parallel
R = ki + 4j N is parallel to 3i + 4j. Find k.
Hint
Use the j component to determine λ first.
Worked solution
4 = 4λ gives λ = 1, then k = 3λ = 3.
04 · Opposite sense
R = ki − 6j N is parallel to i + 2j. Find k and state whether the direction is the same.
Hint
The j component sets the sign of λ.
Worked solution
−6 = 2λ gives λ = −3 and k = −3. It points in the opposite direction.
05 · Impossible direction
Can the vector in question 4 point in the same direction as i + 2j?
Hint
The same direction needs positive λ.
Worked solution
No. Its j component forces λ = −3, regardless of k.
04 / Build a perpendicular direction by swapping and changing one sign
This avoids dividing by a component that might be zero.
A perpendicular to ai + bj is −bi + aj.
For nonzero vectors in the plane, rotate the direction through 90°. For example, a perpendicular to 2i + j is −i + 2j. A scalar multiple of this perpendicular also works. This agrees with the negative reciprocal gradient rule when both gradients exist.
06 · Perpendicular
R = ki + 4j N is perpendicular to 2i + j. Find k.
Hint
Write R = λ(−i + 2j).
Worked solution
4 = 2λ gives λ = 2, so k = −2.
07 · Horizontal reference
A nonzero resultant (p − 2)i + 5j N is perpendicular to i. Find p.
Hint
A perpendicular to a horizontal vector is vertical.
Worked solution
p − 2 = 0 gives p = 2. The remaining vector 5j is nonzero and vertical.
05 / Keep both roots of a squared equation initially
The same length can occur on opposite sides of an axis.
R = ki + 4j N has magnitude 5 N.Worked example
k² + 4² = 5²
Apply Pythagoras.
k² = 9, so k = −3 or k = 3
Both give magnitude 5 N.
A northeast direction would select k = 3
Direction is extra information, not a consequence of magnitude alone.
Watch: two endpoints share the same resultant length
Pause, replay or seek freely. The notes explain the same idea and stay in view.
08 · Two roots
Find k if R = 6i + kj N has magnitude 10 N.
Hint
Square the magnitude and keep both square roots.
Worked solution
36 + k² = 100 gives k = ±8.
09 · Select a direction
For question 8, find k if the resultant points southeast.
Hint
Its northward component must be negative.
Worked solution
k = −8. Both roots fit the length, but only the negative root points southeast.
10 · Shifted parameter
Find p if R = (p − 1)i + 12j N has magnitude 13 N.
Hint
Solve for p − 1 before solving for p.
Worked solution
(p − 1)² = 25, so p − 1 = ±5 and p = 6 or p = −4.
06 / Check whether the condition is possible
A vector length cannot be smaller than an absolute component.
11 · No real solution
Can R = ki + 4j N have magnitude 3 N for real k?
Hint
Its vertical component already has magnitude 4 N.
Worked solution
No. k² + 16 = 9 gives k² = −7, which has no real solution.
12 · A minimum magnitude
Find the smallest possible magnitude of ki + 4j N and the corresponding k.
Hint
k² is never negative.
Worked solution
The minimum is √16 = 4 N, attained at k = 0.
13 · Zero is not a direction
R = (p − 2)i N must point east. State the condition on p.
Hint
A positive i component is required.
Worked solution
p > 2. At p = 2 the resultant is zero with no direction; p < 2 points west.
07 / Substitute into every original condition
A root of one equation may fail the rest of the problem.
14 · Combined restrictions
R = ki + 4j N has magnitude 5 N and is perpendicular to 4i − 3j. Find k.
Hint
Find the magnitude roots, then test the perpendicular condition.
Worked solution
Magnitude gives k = ±3. A perpendicular to 4i − 3j is 3i + 4j, so R must have k = 3. The root −3 has the right length but the wrong direction.
A parameter used as a signed component can be negative. A parameter explicitly defined as a mass or force magnitude must meet its physical constraints. Do not discard a negative parameter merely because the context involves forces.
08 / Write the condition, solve, then test the roots
Keep the physical interpretation attached to the algebra.
For equilibrium, set both sums to zero. For direction, use a nonzero scalar multiple with the required sign. For a length, square the components and retain all real roots until the other conditions select them.
Section 1 of 8 · Translate the wording into a precise condition