Moving a force, forces as vectors and practice
Answers are for parents and teachers.
1
Two like forces of 16 N at A and 24 N at B act on a light rod with AB = 50 cm. Find R and AC. The 16 N force then slides 10 cm towards B. Find the new AC.
R = 40 N; 40 × AC = 24 × 50 gives AC = 30 cm. The moving force is 16 N: CC′ = 16 × 10 / 40 = 4 cm towards B, so the new AC = 34 cm. Check with moments about A: 16 × 10 + 24 × 50 = 1360 and 1360 / 40 = 34.
2
A force of 18 N acts at A and a force F at B, both the same way. When F slides 6 cm along the rod, the resultant moves 4 cm. Find F.
6F / (18 + F) = 4 gives 6F = 72 + 4F, so F = 36 N.
3
F₁ = (2, −1) acts at A(−3, 4) and F₂ = (6, −3) at B(5, 0). Find the resultant and the point where its line cuts AB.
F₂ = 3F₁, like, k = 3. R = (8, −4). C = (A + 3B) / 4 = ((−3 + 15) / 4, (4 + 0) / 4) = (3, 1).
4
F₁ = (−1, 3) acts at A(2, 1) and F₂ = (3, −9) at B(4, 5). Find R and the point where its line cuts the line AB. Is that point inside AB?
F₂ = −3F₁, unlike, k = −3. R = (2, −6). C = (A − 3B) / (−2) = (3B − A) / 2 = ((12 − 2) / 2, (15 − 1) / 2) = (5, 7). No: C is outside AB, beyond B, on the side of the larger force.
5
Weights of 2 N, 8 N, 4 N and 6 N hang from the marks 0, 5, 15 and 20 cm of a light rod, as in the figure. Where must a single string be tied to hang the rod level, and what force must it pull with?
The string pulls up with the sum of the weights, 2 + 8 + 4 + 6 = 20 N, at (2 × 0 + 8 × 5 + 4 × 15 + 6 × 20) / 20 = 220 / 20 = 11 cm.