\begin{tabular}{l} Two forces act on an object. The first is a horizontal So N force. The \ second is a 40 N force acting at an angle to the horizontal. \ Use a graphical method drawn to a scale of $$ 1 \mathrm{~mm}: 1 \mathrm{~N} $$ to find the \ resultant if the 40 N force Is Inclined at: \ $$ 1.1 .1 ~ $$ \ \hline $$ 0^{\circ} $$ to the horizontal \ $$ 1.1 .260^{\circ} $$ to the horizontal. \ 1.2 Compare the resultant forces and make a generallsation after answering \ the following question. Does the magnitude of the resultant of two \ vectors increase, stay the same or decrease when the angle between the \ vectors increases? \ 1.3 Consider and then state the angle (in degrees) between two vectors for \end{tabular}

Question

UpStudy AI · Accepted Answer

Certainly! Let's tackle each part of your question step-by-step.

---

## **1.1 Graphical Method to Find the Resultant**

Given:
- **Force $$ F_1 $$**: 50 N (Horizontal)
- **Force $$ F_2 $$**: 40 N at an angle $$ 	heta $$ to the horizontal

**Scale Used**: $$ 1 \, 	ext{mm} : 1 \, 	ext{N} $$

Since I can't provide actual drawings, I'll illustrate the graphical method using analytical calculations, which mirror the graphical approach by breaking vectors into components.

### **a) When $$ 	heta = 0^\circ $$ (Force $$ F_2 $$ is Horizontal)**

1. **Components of $$ F_2 $$**:
   - $$ F_{2x} = F_2 \cos 	heta = 40 \cos 0^\circ = 40 	imes 1 = 40 \, 	ext{N} $$
   - $$ F_{2y} = F_2 \sin 	heta = 40 \sin 0^\circ = 40 	imes 0 = 0 \, 	ext{N} $$

2. **Resultant Components**:
   - $$ R_x = F_{1} + F_{2x} = 50 + 40 = 90 \, 	ext{N} $$
   - $$ R_y = F_{2y} = 0 \, 	ext{N} $$

3. **Magnitude and Direction of Resultant $$ R $$**:
   - $$ R = \sqrt{R_x^2 + R_y^2} = \sqrt{90^2 + 0^2} = 90 \, 	ext{N} $$
   - **Direction**: $$ 0^\circ $$ (same as the horizontal)

**_Graphical Interpretation_**: In this case, both forces are aligned horizontally. The resultant is simply the sum of the two forces in the horizontal direction.

### **b) When $$ 	heta = 60^\circ $$ (Force $$ F_2 $$ is Inclined at $$ 60^\circ $$)**

1. **Components of $$ F_2 $$**:
   - $$ F_{2x} = 40 \cos 60^\circ = 40 	imes 0.5 = 20 \, 	ext{N} $$
   - $$ F_{2y} = 40 \sin 60^\circ \approx 40 	imes 0.8660 = 34.64 \, 	ext{N} $$

2. **Resultant Components**:
   - $$ R_x = 50 + 20 = 70 \, 	ext{N} $$
   - $$ R_y = 34.64 \, 	ext{N} $$

3. **Magnitude and Direction of Resultant $$ R $$**:
   - $$ R = \sqrt{70^2 + 34.64^2} \approx \sqrt{4900 + 1197} \approx \sqrt{6097} \approx 78.1 \, 	ext{N} $$
   - **Direction**: $$ \alpha = \arctan\left(\frac{R_y}{R_x}ight) = \arctan\left(\frac{34.64}{70}ight) \approx 26.3^\circ $$ above the horizontal

**_Graphical Interpretation_**: Here, $$ F_2 $$ has both horizontal and vertical components. The resultant vector $$ R $$ is found by adding the horizontal and vertical components separately, resulting in a vector that is less than the sum of the magnitudes due to the angle.

---

## **1.2 Comparing Resultant Forces and Generalization**

### **Comparison:**

- **At $$ 0^\circ $$**:
  - **Resultant $$ R $$**: 90 N horizontally
- **At $$ 60^\circ $$**:
  - **Resultant $$ R $$**: Approximately 78.1 N at $$ 26.3^\circ $$ above the horizontal

**Observation**: When the angle $$ 	heta $$ increases from $$ 0^\circ $$ to $$ 60^\circ $$, the magnitude of the resultant force **decreases**.

### **Generalization:**

**Question**: Does the magnitude of the resultant of two vectors increase, stay the same, or decrease when the angle between the vectors increases?

**Answer**: **The magnitude of the resultant decreases as the angle between the two vectors increases.**

**Explanation**:
- **Maximum Resultant**: When the two vectors are aligned in the same direction ($$ 	heta = 0^\circ $$). The resultant is simply the sum of the magnitudes.
- **Minimum Resultant**: When the two vectors are aligned in opposite directions ($$ 	heta = 180^\circ $$). The resultant is the difference of the magnitudes.
- **Intermediate Angles**: For angles between $$ 0^\circ $$ and $$ 180^\circ $$, the resultant magnitude decreases as the angle increases from $$ 0^\circ $$ to $$ 180^\circ $$.

---

## **1.3 Angle Between Two Vectors for Specific Resultant Conditions**

**Question**: *Consider and then state the angle (in degrees) between two vectors when...*  
*(Note: The question appears incomplete. I'll assume it seeks the angle for maximum or minimum resultant.)*

### **a) Angle for Maximum Resultant**

**Answer**: **$$ 0^\circ $$**

**Explanation**: When two vectors are aligned in the same direction, the resultant is maximized. The magnitude of the resultant is the sum of the magnitudes of the individual vectors.

### **b) Angle for Minimum Resultant**

**Answer**: **$$ 180^\circ $$**

**Explanation**: When two vectors are aligned in opposite directions, the resultant is minimized. The magnitude of the resultant is the absolute difference of the magnitudes of the individual vectors.

### **c) Angle Where Resultant is Zero**

**Answer**: **Only possible if vectors are equal in magnitude and aligned at $$ 180^\circ $$**

**Explanation**: If two vectors have equal magnitudes and are directed opposite to each other, their resultant will be zero.

---

## **Summary**

- **When the angle between two vectors increases**, assuming one vector's direction is fixed, **the magnitude of the resultant decreases**.
- **Maximum resultant** occurs at $$ 0^\circ $$ (same direction).
- **Minimum resultant** occurs at $$ 180^\circ $$ (opposite directions).
- **Graphical methods** involve breaking vectors into components, adding corresponding components, and then finding the magnitude and direction of the resultant vector.

---

If you have any further questions or need clarification on a specific part, feel free to ask!
When two forces act on an object, their resultant force decreases as the angle between them increases. At $$0^\circ$$, the resultant is the sum of the forces, and at $$180^\circ$$, it's the difference.

Solución de tutoría real

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Physics Questions