Math Problem Statement

Diketahui besar gaya 1 dan 2 adalah: F1 = 10 N; F2 = 8 N. Tentukan: (a) Besar resultan vektor, (b) Arah Resultan vektor, (c) Gambarkan resultan dari kedua vektor tersebut!

Solution

Let's solve this step-by-step:

Given:

  • F1=10NF_1 = 10 \, \text{N} at an angle of 4545^\circ from the positive YY-axis.
  • F2=8NF_2 = 8 \, \text{N} at an angle of 3030^\circ from the negative XX-axis.

We need to find:

  1. The magnitude of the resultant vector (RR).
  2. The direction of the resultant vector (θ\theta from a specified axis).
  3. A diagram of the resultant vector.

Step 1: Resolve F1F_1 and F2F_2 into their components

For F1F_1:

  • F1x=F1sin(45)=10sin(45)=1022=7.07NF_{1x} = F_1 \cdot \sin(45^\circ) = 10 \cdot \sin(45^\circ) = 10 \cdot \frac{\sqrt{2}}{2} = 7.07 \, \text{N} (positive along the X-axis)
  • F1y=F1cos(45)=10cos(45)=1022=7.07NF_{1y} = F_1 \cdot \cos(45^\circ) = 10 \cdot \cos(45^\circ) = 10 \cdot \frac{\sqrt{2}}{2} = 7.07 \, \text{N} (positive along the Y-axis)

For F2F_2:

  • F2x=F2cos(30)=8cos(30)=832=6.93NF_{2x} = F_2 \cdot \cos(30^\circ) = 8 \cdot \cos(30^\circ) = 8 \cdot \frac{\sqrt{3}}{2} = 6.93 \, \text{N} (negative along the X-axis)
  • F2y=F2sin(30)=8sin(30)=812=4NF_{2y} = F_2 \cdot \sin(30^\circ) = 8 \cdot \sin(30^\circ) = 8 \cdot \frac{1}{2} = 4 \, \text{N} (negative along the Y-axis)

Step 2: Find the resultant components RxR_x and RyR_y

  • Rx=F1x+F2x=7.076.93=0.14NR_x = F_{1x} + F_{2x} = 7.07 - 6.93 = 0.14 \, \text{N}
  • Ry=F1y+F2y=7.074=3.07NR_y = F_{1y} + F_{2y} = 7.07 - 4 = 3.07 \, \text{N}

Step 3: Calculate the magnitude of the resultant vector RR

R=(Rx)2+(Ry)2=(0.14)2+(3.07)2=0.0196+9.4249=9.44453.07NR = \sqrt{(R_x)^2 + (R_y)^2} = \sqrt{(0.14)^2 + (3.07)^2} = \sqrt{0.0196 + 9.4249} = \sqrt{9.4445} \approx 3.07 \, \text{N}

Step 4: Determine the direction of the resultant vector θ\theta

θ=arctan(RyRx)=arctan(3.070.14)87.39\theta = \arctan\left(\frac{R_y}{R_x}\right) = \arctan\left(\frac{3.07}{0.14}\right) \approx 87.39^\circ

This angle is measured from the positive XX-axis towards the YY-axis.

Answers:

  1. Magnitude of the resultant vector: R3.07NR \approx 3.07 \, \text{N}
  2. Direction of the resultant vector: θ87.39\theta \approx 87.39^\circ from the positive XX-axis.
  3. Diagram: I can describe it, or you can draw it based on the components RxR_x and RyR_y with the resultant angle.

Would you like further details on any of these steps or assistance with the diagram?


Here are some related questions to explore further:

  1. How would the result change if the angles were measured from a different reference axis?
  2. What would happen to the resultant vector if F1F_1 and F2F_2 had the same magnitude?
  3. How does changing the angle of F1F_1 affect the magnitude and direction of the resultant vector?
  4. What is the physical interpretation of vector addition in terms of force equilibrium?
  5. How can we use vector components to simplify complex force systems in physics?

Tip: When working with vectors, resolving into components and using trigonometric identities simplifies calculations greatly.

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Trigonometry
Pythagorean Theorem
Vector Resolution

Formulas

Component form: F_x = F * cos(θ), F_y = F * sin(θ)
Resultant magnitude: R = sqrt((R_x)^2 + (R_y)^2)
Angle of resultant: θ = arctan(R_y / R_x)

Theorems

Pythagorean Theorem
Trigonometric Identities

Suitable Grade Level

Grades 10-12