Question

iv. solving resultant vectors

1. same direction: two forces act on a box: 15 n east and 25 n east. find the resultant force.
2. opposite direction: one force pulls a rope with 60 n east, and another pulls with 40 n west. find the resultant force and its direction.
3. perpendicular directions (right angle): a boat moves 6 m north and 8 m east. find the resultant displacement. (hint: use r = √a² + b²)
4. finding magnitude and direction: a plane travels 12 km north and 5 km east. find the resultant and its direction angle from the north. (hint: r = √a² + b², tanθ = b/a)
5. challenging: a student walks 4 m east, then 3 m south, and then 2 m west. find the resultant displacement from the starting point.

Explanation:

Step1: Analyze the first problem (Same Direction)

When vectors are in the same direction, we add their magnitudes.
Let \(F_1 = 15N\) and \(F_2=25N\). The formula for the resultant force \(F_R\) is \(F_R=F_1 + F_2\).
Substitute the values: \(F_R=15 + 25\).

Step2: Analyze the second problem (Opposite Direction)

Let the east - direction be positive. So \(F_1 = 60N\) (east) and \(F_2=- 40N\) (west). The formula for the resultant force \(F_R\) is \(F_R=F_1+F_2\).
Substitute the values: \(F_R = 60+( - 40)=20N\). Since the result is positive, the direction is east.

Step3: Analyze the third problem (Perpendicular Directions)

Let \(a = 6m\) (north) and \(b = 8m\) (east). Using the Pythagorean theorem \(R=\sqrt{a^{2}+b^{2}}\).
Substitute \(a = 6\) and \(b = 8\): \(R=\sqrt{6^{2}+8^{2}}=\sqrt{36 + 64}=\sqrt{100}=10m\).

Step4: Analyze the fourth problem (Magnitude and Direction)

Let \(a = 12km\) (north) and \(b = 5km\) (east).
For the magnitude, use \(R=\sqrt{a^{2}+b^{2}}\). Substitute \(a = 12\) and \(b = 5\): \(R=\sqrt{12^{2}+5^{2}}=\sqrt{144 + 25}=\sqrt{169}=13km\).
For the direction \(\theta\) (from the north), use \(\tan\theta=\frac{b}{a}\). Substitute \(a = 12\) and \(b = 5\): \(\tan\theta=\frac{5}{12}\), so \(\theta=\arctan(\frac{5}{12})\approx22.6^{\circ}\) east of north.

Step5: Analyze the fifth problem (Challenging)

In the x - direction (east - west): The student moves \(4m\) east and \(2m\) west. The net displacement in the x - direction \(x=4 - 2=2m\) east.
In the y - direction (north - south): The student moves \(3m\) south. Let \(x = 2m\) and \(y=-3m\).
Using the Pythagorean theorem \(R=\sqrt{x^{2}+y^{2}}\). Substitute \(x = 2\) and \(y=-3\): \(R=\sqrt{2^{2}+(-3)^{2}}=\sqrt{4 + 9}=\sqrt{13}\approx3.61m\).

Answer:

1. \(40N\) east
2. \(20N\) east
3. \(10m\)
4. \(13km\), approximately \(22.6^{\circ}\) east of north
5. Approximately \(3.61m\)