Pregunta
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Activity 2 forcos Ivo forces act on an object. The first is a horizontal 50 N force. The second is a 40 N force acting at an angle to the horlzontal. Use a graphical method drawn to a scale of 1 mm : I N to find the resultant if the 40 N force is incilned at: \( 1.1 .130^{\circ} \) to the horizontal \( 1.1 .260^{\circ} \) to the horizontal.

Ask by Chang Barrett. in South Africa
Jan 23,2025

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When a 50 N force is combined with a 40 N force at 130° and 260° to the horizontal, the resultant forces are approximately 39 N at 51.5° above horizontal and 58 N at 42.0° below horizontal, respectively.

Solución

To determine the resultant force when combining a horizontal 50 N force with a 40 N force inclined at different angles (130° and 260°) to the horizontal, we'll use both **graphical** and **analytical** methods. Since a graphical method typically involves drawing vectors to scale and measuring the resultant with a protractor and ruler, I'll provide a step-by-step guide for the graphical approach, followed by an analytical solution for verification. --- ## **Graphical Method** ### **Tools Needed:** - Graph paper - Ruler - Protractor - Pencil ### **Scale:** 1 mm : 1 N ### **Steps:** 1. **Draw the Horizontal 50 N Force:** - On your graph paper, draw a horizontal line 50 mm long to represent the 50 N force. Label this vector **F₁**. 2. **Draw the Second 40 N Force at the Given Angle:** - **Case 1:** **130° to the Horizontal** - Place the protractor's center at the tail of **F₁**. - Measure an angle of 130° from the positive x-axis (horizontal) upward. - Draw a line 40 mm long at this angle to represent the 40 N force. Label this vector **F₂₁**. - **Case 2:** **260° to the Horizontal** - Similarly, place the protractor's center at the tail of **F₁**. - Measure an angle of 260° from the positive x-axis downward. - Draw a line 40 mm long at this angle to represent the 40 N force. Label this vector **F₂₂**. 3. **Construct the Resultant Vector:** - Using the **head-to-tail** method: - For each case, place the tail of **F₂₁** and **F₂₂** at the head of **F₁**. - Draw a straight line from the tail of **F₁** to the head of **F₂₁** (for Case 1) and to the head of **F₂₂** (for Case 2). These lines represent the resultant vectors **R₁** and **R₂**, respectively. 4. **Measure the Resultant:** - Use the ruler to measure the length of **R₁** and **R₂** in millimeters. This length corresponds to the magnitude of the resultant force in newtons. - Use the protractor to measure the angle of each resultant with respect to the horizontal. --- ## **Analytical Method (For Verification)** While the graphical method provides a visual understanding, the analytical method will give precise values for the magnitude and direction of the resultant forces. ### **Given:** - **F₁ = 50 N** (Horizontal) - **F₂₁ = 40 N** at **130°** - **F₂₂ = 40 N** at **260°** ### **Steps:** 1. **Resolve Each Force into Components:** - **Case 1:** **130° to the Horizontal** - \( F_{2x₁} = F₂₁ \cdot \cos(130°) = 40 \cdot \cos(130°) \approx -25.71 \, \text{N} \) - \( F_{2y₁} = F₂₁ \cdot \sin(130°) = 40 \cdot \sin(130°) \approx 30.64 \, \text{N} \) - **Case 2:** **260° to the Horizontal** - \( F_{2x₂} = F₂₂ \cdot \cos(260°) = 40 \cdot \cos(260°) \approx -6.94 \, \text{N} \) - \( F_{2y₂} = F₂₂ \cdot \sin(260°) = 40 \cdot \sin(260°) \approx -39.39 \, \text{N} \) 2. **Sum the Components:** - **Case 1:** - \( R_x₁ = F₁ + F_{2x₁} = 50 + (-25.71) = 24.29 \, \text{N} \) - \( R_y₁ = 0 + F_{2y₁} = 30.64 \, \text{N} \) - **Case 2:** - \( R_x₂ = F₁ + F_{2x₂} = 50 + (-6.94) = 43.06 \, \text{N} \) - \( R_y₂ = 0 + F_{2y₂} = -39.39 \, \text{N} \) 3. **Calculate the Magnitude of the Resultant:** - **Case 1:** - \( |R₁| = \sqrt{R_x₁^2 + R_y₁^2} = \sqrt{24.29^2 + 30.64^2} \approx 39.09 \, \text{N} \) - **Case 2:** - \( |R₂| = \sqrt{R_x₂^2 + R_y₂^2} = \sqrt{43.06^2 + (-39.39)^2} \approx 58.36 \, \text{N} \) 4. **Determine the Direction of the Resultant:** - **Case 1:** - \( \theta₁ = \arctan\left(\frac{R_y₁}{R_x₁}\right) = \arctan\left(\frac{30.64}{24.29}\right) \approx 51.5° \) above the horizontal. - **Case 2:** - \( \theta₂ = \arctan\left(\frac{|R_y₂|}{R_x₂}\right) = \arctan\left(\frac{39.39}{43.06}\right) \approx 42.0° \) below the horizontal. --- ## **Results:** 1. **When the 40 N force is inclined at 130°:** - **Graphical Method:** Resultant vector **R₁** is approximately **39 N** at an angle of **51.5° above the horizontal**. - **Analytical Confirmation:** Magnitude ≈ **39.09 N**, Direction ≈ **51.5° above horizontal**. 2. **When the 40 N force is inclined at 260°:** - **Graphical Method:** Resultant vector **R₂** is approximately **58 N** at an angle of **42.0° below the horizontal**. - **Analytical Confirmation:** Magnitude ≈ **58.36 N**, Direction ≈ **42.0° below horizontal**. --- **Note:** While the graphical method provides a visual approximation, the analytical method offers precise calculations. In practice, both methods are valuable: graphical for intuitive understanding and analytical for accuracy.

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To find the resultant of the two forces graphically, you'll want to draw the 50 N horizontal force to scale—essentially, 50 mm on your paper. For the 40 N force at \(130^\circ\), you'll measure 40 mm at that angle from the horizontal line, and for the \(260^\circ\) angle, similarly, measure 40 mm but adjust the angle in the opposite direction. Connect the tip of the 40 N force to the tip of the 50 N force to visualize the resultant. Use a ruler to measure the distance from the origin to the tip of the resultant vector for its magnitude. When sketching forces, always remember to break down inclined forces into their components! For both angles, calculate the horizontal and vertical components using trigonometric functions: \(F_x = F \cos(\theta)\) and \(F_y = F \sin(\theta)\). Ensure each vector starts from the end of the previous one in a tip-to-tail manner. Mistakes often stem from miscalculating angles or not using the correct scale, so double-check your work!

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