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

### Given:
- **Force by Yui** \( F_Y = 120 \, \text{N} \) at an angle \( \theta = 36.0^\circ \) from the positive \( x \)-axis.
- **Force by Gahama** \( F_G = 150 \, \text{N} \) along the negative \( x \)-axis.

### Part A: Magnitude and Direction of the Net Force

#### Step 1: Resolve Yui's force \( F_Y \) into components:
- \( F_{Yx} = F_Y \cos \theta = 120 \times \cos(36^\circ) \)
- \( F_{Yy} = F_Y \sin \theta = 120 \times \sin(36^\circ) \)

#### Step 2: Gahama's force is entirely along the \( -x \)-axis, so:
- \( F_{Gx} = -150 \, \text{N} \)
- \( F_{Gy} = 0 \, \text{N} \)

#### Step 3: Calculate the net force in both \( x \)- and \( y \)-directions:
- Net \( x \)-component: \( F_x = F_{Yx} + F_{Gx} \)
- Net \( y \)-component: \( F_y = F_{Yy} + F_{Gy} \)

#### Step 4: Magnitude of the net force:
\[
F_{\text{net}} = \sqrt{F_x^2 + F_y^2}
\]

#### Step 5: Direction of the net force:
\[
\theta_{\text{net}} = \tan^{-1}\left(\frac{F_y}{F_x}\right)
\]

### 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 \( F_I \) should equal \( F_{\text{net}} \), but in the opposite direction (same magnitude, \( 180^\circ \) 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 \( x \)-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:** \( 126.88^\circ + 180^\circ = 306.88^\circ \) from the positive \( x \)-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 \( -x \)-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.

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?

Learn how to calculate the net force when two forces act on a circular card at different angles. This problem involves vector addition, trigonometry, and the equilibrium of forces. Suitable for high school students in Grades 10-12.

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