Math Problem Statement

Determine the (a) magnitude and (b) direction for the resultant that occurs when the vectors A = 16.0 m, B = 10.0 m, C = 11.0 m, and D = 29.0 m are added together. Specify the direction as a positive (counterclockwise) angle from the +x axis.

Solution

The problem in the image involves adding four displacement vectors, labeled AA, BB, CC, and DD, and finding the magnitude and direction of the resultant vector. The angles and magnitudes of the vectors are given as follows:

  • A=16.0mA = 16.0 \, \text{m} at 3535^\circ above the positive xx-axis.
  • B=10.0mB = 10.0 \, \text{m} along the positive yy-axis (9090^\circ).
  • C=11.0mC = 11.0 \, \text{m} at 3535^\circ below the negative xx-axis.
  • D=29.0mD = 29.0 \, \text{m} at 5050^\circ below the positive xx-axis.

Solution Steps:

To find the resultant vector, we need to calculate the sum of the xx-components and yy-components of each vector. Then, we use the Pythagorean theorem to find the magnitude of the resultant vector, and the inverse tangent to find its direction.

1. Resolve the vectors into components:

  • Ax=Acos(35)=16.0cos(35)A_x = A \cos(35^\circ) = 16.0 \cos(35^\circ)

  • Ay=Asin(35)=16.0sin(35)A_y = A \sin(35^\circ) = 16.0 \sin(35^\circ)

  • Bx=0B_x = 0 (since BB is along the yy-axis)

  • By=B=10.0mB_y = B = 10.0 \, \text{m}

  • Cx=Ccos(35)=11.0cos(35)C_x = C \cos(35^\circ) = -11.0 \cos(35^\circ) (negative since it's to the left of the origin)

  • Cy=Csin(35)=11.0sin(35)C_y = C \sin(35^\circ) = -11.0 \sin(35^\circ) (negative since it's downward)

  • Dx=Dcos(50)=29.0cos(50)D_x = D \cos(50^\circ) = 29.0 \cos(50^\circ)

  • Dy=Dsin(50)=29.0sin(50)D_y = D \sin(50^\circ) = -29.0 \sin(50^\circ) (negative since it's downward)

2. Sum the components:

  • Rx=Ax+Bx+Cx+DxR_x = A_x + B_x + C_x + D_x
  • Ry=Ay+By+Cy+DyR_y = A_y + B_y + C_y + D_y

3. Find the magnitude of the resultant vector:

R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}

4. Find the direction of the resultant vector:

θ=tan1(RyRx)\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right)

I will now calculate these values.The magnitude and direction of the resultant vector are as follows:

  • Magnitude: 24.58m24.58 \, \text{m}
  • Direction: 22.35-22.35^\circ (measured counterclockwise from the positive xx-axis, so this is 22.35° below the positive xx-axis).

Would you like further clarification or details on any of the steps? Here are five related questions you can explore:

  1. How do we add vectors graphically versus analytically?
  2. How would the results change if one of the vectors had a different direction?
  3. What is the significance of breaking vectors into components?
  4. How does the resultant vector help in understanding physical motion?
  5. How can we interpret negative angles in vector direction?

Tip: Always break vectors into components to simplify problems involving vector addition!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector Addition
Trigonometry
Vector Components
Pythagorean Theorem

Formulas

Ax = A * cos(θ)
Ay = A * sin(θ)
Rx = Ax + Bx + Cx + Dx
Ry = Ay + By + Cy + Dy
Resultant Magnitude: R = sqrt(Rx^2 + Ry^2)
Resultant Angle: θ = tan^(-1)(Ry / Rx)

Theorems

Pythagorean Theorem
Trigonometric Ratios

Suitable Grade Level

Grades 11-12 (Advanced High School Physics)