Math Problem Statement

Yui and Gahama are tugging on a circular card. Yui pulls on the card with a force of 120 N at an angle of 36.0° from the positive x-axis, while Gahama pulls with a force of 150 N along the -x-axis. What is the magnitude and direction of the net force on the card? Iroha joins in and tugs on the card so that the net force vanishes. What is the magnitude and direction of Iroha's force?

Solution

Let's break down the problem and solve it step by step.

Given:

  • Force by Yui FY=120NF_Y = 120 \, \text{N} at an angle θ=36.0\theta = 36.0^\circ from the positive xx-axis.
  • Force by Gahama FG=150NF_G = 150 \, \text{N} along the negative xx-axis.

Part A: Magnitude and Direction of the Net Force

Step 1: Resolve Yui's force FYF_Y into components:

  • FYx=FYcosθ=120×cos(36)F_{Yx} = F_Y \cos \theta = 120 \times \cos(36^\circ)
  • FYy=FYsinθ=120×sin(36)F_{Yy} = F_Y \sin \theta = 120 \times \sin(36^\circ)

Step 2: Gahama's force is entirely along the x-x-axis, so:

  • FGx=150NF_{Gx} = -150 \, \text{N}
  • FGy=0NF_{Gy} = 0 \, \text{N}

Step 3: Calculate the net force in both xx- and yy-directions:

  • Net xx-component: Fx=FYx+FGxF_x = F_{Yx} + F_{Gx}
  • Net yy-component: Fy=FYy+FGyF_y = F_{Yy} + F_{Gy}

Step 4: Magnitude of the net force:

Fnet=Fx2+Fy2F_{\text{net}} = \sqrt{F_x^2 + F_y^2}

Step 5: Direction of the net force:

θnet=tan1(FyFx)\theta_{\text{net}} = \tan^{-1}\left(\frac{F_y}{F_x}\right)

Part B: Iroha's Force to Cancel the Net Force

To make the net force vanish, Iroha's force must exactly cancel the net force from Part A. So, the magnitude of Iroha's force FIF_I should equal FnetF_{\text{net}}, but in the opposite direction (same magnitude, 180180^\circ opposite angle).

Let me calculate these values for you.### Part A: The magnitude of the net force on the card is approximately 88.18 N, and its direction is about 126.88° from the positive xx-axis (measured counterclockwise).

Part B:

For the net force to vanish, Iroha must exert a force of the same magnitude but in the opposite direction. Therefore, Iroha's force should have:

  • Magnitude: 88.18 N
  • Direction: 126.88+180=306.88126.88^\circ + 180^\circ = 306.88^\circ from the positive xx-axis (or equivalently, 126.88° in the opposite direction).

Would you like more details on any part of the solution, or do you have any further questions?

Here are some related questions to expand your understanding:

  1. How would the result change if Yui's angle was different?
  2. What if Gahama applied his force at an angle instead of along the x-x-axis?
  3. How does the direction of Iroha's force affect the system's equilibrium?
  4. How can vector components simplify the analysis of forces in multiple dimensions?
  5. How would this scenario differ if we added friction or other external forces?

Tip: When dealing with forces, always break them into components to analyze their effects in each direction independently.

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Math Problem Analysis

Mathematical Concepts

Vector addition
Trigonometry
Force equilibrium

Formulas

F_x = F_{Yx} + F_{Gx}
F_y = F_{Yy} + F_{Gy}
F_{net} = sqrt(F_x^2 + F_y^2)
θ_{net} = tan^{-1}(F_y / F_x)

Theorems

Pythagorean theorem
Trigonometric identities

Suitable Grade Level

Grades 10-12