Application of Vectors
Vectors in Solving Simple Problems on Velocities, Displacements and Forces
Vector knowledge is applicable in solving many practical problems as in the following examples.
A student walks 40 m in the direction S 450 E from the dormitory to the parade ground and then he walks 100m due east to his classroom. Find his displacement from dormitory to the classroom.
Solution
Consider the following figure describing the displacement which joins the dormitory D. parade ground P and Classroom C.
∣ D C ⃗ ∣ 2 = 400 2 + 100 2 − 2 ( 400 ) ( 100 ) cos 45 ∘ |\vec{DC}|^2 = 400^2 + 100^2 - 2(400)(100) \cos 45^\circ ∣ D C ∣ 2 = 40 0 2 + 10 0 2 − 2 ( 400 ) ( 100 ) cos 4 5 ∘
= 160000 + 10000 − 80000 × cos 45 ∘ = 160000 + 10000 - 80000 \times \cos 45^\circ = 160000 + 10000 − 80000 × cos 4 5 ∘ = 170000 − 80000 × 0.707 = 170000 - 80000 \times 0.707 = 170000 − 80000 × 0.707 = 170000 − 56560 = 170000 - 56560 = 170000 − 56560 = 113440 = 113440 = 113440 ∣ D C ⃗ ∣ = 113440 ≈ 336.8 m |\vec{DC}| = \sqrt{113440} \approx 336.8 \, \text{m} ∣ D C ∣ = 113440 ≈ 336.8 m
Let ∠ C D P = θ \angle CDP = \theta ∠ C D P = θ
sin θ 100 = sin 45 ∘ 336.8 \frac{\sin \theta}{100} = \frac{\sin 45^\circ}{336.8} 100 sin θ = 336.8 sin 4 5 ∘ Thus,
sin θ = 100 × sin 45 ∘ 336.8 \sin \theta = \frac{100 \times \sin 45^\circ}{336.8} sin θ = 336.8 100 × sin 4 5 ∘ = 100 × 0.707 336.8 = \frac{100 \times 0.707}{336.8} = 336.8 100 × 0.707 = 70.7 336.8 = \frac{70.7}{336.8} = 336.8 70.7 sin θ ≈ 0.21 \sin \theta \approx 0.21 sin θ ≈ 0.21 Therefore,
θ ≈ 12 ∘ \theta \approx 12^\circ θ ≈ 1 2 ∘ Hence, the bearing is:
S ( 45 ∘ − 12 ∘ ) E = S 33 ∘ E S(45^\circ - 12^\circ)E = S 33^\circ E S ( 4 5 ∘ − 1 2 ∘ ) E = S 3 3 ∘ E ∴ The boys’ displacement from the dormitory to the classroom is 336.8 m at a bearing of S 33 ∘ E . \therefore \text{The boys' displacement from the dormitory to the classroom is } 336.8 \, \text{m at a bearing of } S 33^\circ E. ∴ The boys’ displacement from the dormitory to the classroom is 336.8 m at a bearing of S 3 3 ∘ E . Example 2
Three forces F1 = (3,4), F2 = (5,-2) and F3 = (4,3) measured in Newtons act at point O (0,0)
Determine the magnitude and direction of their resultant.
Calculate the magnitude and direction of the opposite of the resultant force.
Solution
(a) Let F F F
F = F 1 + F 2 + F 3 F = F_1 + F_2 + F_3 F = F 1 + F 2 + F 3
= ( 3 , 4 ) + ( 5 , − 2 ) + ( 4 , 3 ) = (3, 4) + (5, -2) + (4, 3) = ( 3 , 4 ) + ( 5 , − 2 ) + ( 4 , 3 )
= ( 12 , 5 ) = (12, 5) = ( 12 , 5 )
∣ F ∣ = 13 N |F| = 13 \text{ N} ∣ F ∣ = 13 N
Also cos α = 12 13 = 0.92307 \cos\alpha = \frac{12}{13} = 0.92307 cos α = 13 12 = 0.92307
α = 22.6 ∘ \alpha = 22.6^\circ α = 22. 6 ∘
α = 22.6 ∘ \alpha = 22.6^\circ α = 22. 6 ∘
From the figure above, the bearing of F F F N 67.4 ∘ E \text{N }67.4^\circ \text{E} N 67. 4 ∘ E
Therefore the resultant force is 13 N at the bearing of N 67.4 ∘ E \text{N }67.4^\circ \text{E} N 67. 4 ∘ E
(b) Let the force opposite to F be Fo, then Fo = -F = - (12, 5) = (-12, -5)
Fo = 13N and its bearing is ( 67.4 ∘ + 180 ∘ ) = 247.4 ∘ (67.4^\circ + 180^\circ) = 247.4^\circ ( 67. 4 ∘ + 18 0 ∘ ) = 247. 4 ∘
So the magnitude and direction of the force opposite to the resultant force is 13N and S 67.4 ∘ 67.4^\circ 67. 4 ∘
