### Problem DescriptionA machine part consists of a thin, uniform horizontal bar and a vertical bar connected at a hinge:1. **Horizontal Bar**: - Mass: 4.00 kg - Length: 1.50 m - Positioned horizontally with its hinge on the left.2. **Vertical Bar**: - Mass: 3.00 kg - Length: 1.80 m - Positioned vertically with its hinge at the top.3. **Ball**: - Mass: 2.00 kg - Attached to the bottom end of the vertical bar. The objective is to find the center of mass of this system relative to the hinge.### Steps to Solve the Problem**a. Center of Mass of the Top Bar**- **X-component:** Since the bar is horizontal, the center of mass is at \( \frac{1.50 \, \text{m}}{2} = 0.75 \, \text{m} \) from the hinge.- **Y-component:** There is no vertical displacement, so it is \( 0 \, \text{m} \).**b. Center of Mass of the Vertical Bar**- **X-component:** There is no horizontal displacement, so it is \( 0 \, \text{m} \).- **Y-component:** The center of mass is at \( \frac{1.80 \, \text{m}}{2} = 0.90 \, \text{m} \) downward from the hinge.**c. Center of Mass of the Ball**- **X-component:** Since the ball is directly below the hinge along the vertical bar, it is \( 0 \, \text{m} \).- **Y-component:** The ball is \( 1.80 \, \text{m} \) downward from the hinge.**d. Center of Mass of the Entire Machine Part**To find the overall center of mass, calculate the weighted average of the components from parts (a), (b), and (c).- **X-component Calculation:** \[ \text{Total mass} = 4.00 \, \text{kg} + 3.00 \, \text{kg} + 2.00 \, \text{kg} = 9.00 \, \text{kg}

Introduction: Centre of mass of a body is defined as a point where whole mass of a body is considered to be concentrated. It can be treated as balanced point of an object. It is a point where force is applied is get linear acceleration without any angular acceleration. If an object is uniform in nature, like a rod, its center of mass should lie on the centre of rod length.Solution: Let's make FBD of the figure shown above where hinge is taken as origin (0,0) of coordinate system as per FBD. Part (a): Position of center of mass (mt)of top bar relative to hinge: As center of mass of uniform rod should lie on the center of the rod. Distance of COM of top bar from hinge= -0.75 m (in x-axis) and 0 (in y axis) Hence, Coordinate of center of mass of vertical rod is (-0.75, 0). Part (b): Position of center of mass (mv) of vertical bar relative to hinge: As center of mass of uniform rod should lie on the center of the rod. Distance of COM of vertical bar from hinge= 0 (in x-axis) and -0.9 (in y axis) Hence, Coordinate of center of mass of vertical rod is (0, -0.9). Part (c): Position of center of mass (mb) of ball relative to hinge: As center of mass of small ball should lie on the center of the ball. As shown in figure total length in vertical direction is 1.8 m which is from hinge to center of ball. Distance of COM of vertical rod from hinge= 0 (in x-axis) and -1.8 (in y axis) Hence, Coordinate of center of mass of vertical rod is (0, -1.8). Part (d): Position of center of mass of complete machine relative to hinge: Center of mass in X- axis: ξ mx = (mt . xt) + (mv .xv) + (mb . xb)mt + mv + mb= (4 x -0.75) + (3 x 0) + (2 x 0)4 + 3 + 2= -0.33 m. Center of mass in X- axis: ξ mx = (mt . yt) + (mv .yv) + (mb . yb)mt + mv + mb= (4 x 0) + (3 x -0.9) + (2 x -1.8)4 + 3 + 2= -0.7 m. Hence, Coordinate of complete machine relative to hinge= (-0.33, -0.7)

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Physics

4. A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (see figure). -1.50 m- 4.00 kg a. Find the position of the center of mass of the top bar relative to the hinge. Write out its x- and y-components. b. Find the position of the center of mass of the vertical bar relative to the hinge. Write out its x- and y-components. c. Find the center of mass of the ball relative to the hinge. Write out its x- and y-components. d. Use your answers to parts (a)-(c) to find the center of mass of the entire machine part relative to the hinge. Write out its x- and y-components. Hinge 3.00 kg 1.80 m 2.00 kg

4. A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (see figure). -1.50 m- 4.00 kg a. Find the position of the center of mass of the top bar relative to the hinge. Write out its x- and y-components. b. Find the position of the center of mass of the vertical bar relative to the hinge. Write out its x- and y-components. c. Find the center of mass of the ball relative to the hinge. Write out its x- and y-components. d. Use your answers to parts (a)-(c) to find the center of mass of the entire machine part relative to the hinge. Write out its x- and y-components. Hinge 3.00 kg 1.80 m 2.00 kg

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Torque

Torque is the rotational equivalent of linear force in physics and mechanics. Based on the course of study, it is also known as the moment of force, rotational force, or turning effect.

Question

### Problem Description

A machine part consists of a thin, uniform horizontal bar and a vertical bar connected at a hinge:

1. **Horizontal Bar**:
- Mass: 4.00 kg
- Length: 1.50 m
- Positioned horizontally with its hinge on the left.

2. **Vertical Bar**:
- Mass: 3.00 kg
- Length: 1.80 m
- Positioned vertically with its hinge at the top.

3. **Ball**:
- Mass: 2.00 kg
- Attached to the bottom end of the vertical bar.

The objective is to find the center of mass of this system relative to the hinge.

### Steps to Solve the Problem

**a. Center of Mass of the Top Bar**

- **X-component:** Since the bar is horizontal, the center of mass is at \( \frac{1.50 \, \text{m}}{2} = 0.75 \, \text{m} \) from the hinge.

- **Y-component:** There is no vertical displacement, so it is \( 0 \, \text{m} \).

**b. Center of Mass of the Vertical Bar**

- **X-component:** There is no horizontal displacement, so it is \( 0 \, \text{m} \).

- **Y-component:** The center of mass is at \( \frac{1.80 \, \text{m}}{2} = 0.90 \, \text{m} \) downward from the hinge.

**c. Center of Mass of the Ball**

- **X-component:** Since the ball is directly below the hinge along the vertical bar, it is \( 0 \, \text{m} \).

- **Y-component:** The ball is \( 1.80 \, \text{m} \) downward from the hinge.

**d. Center of Mass of the Entire Machine Part**

To find the overall center of mass, calculate the weighted average of the components from parts (a), (b), and (c).

- **X-component Calculation:**

\[
\text{Total mass} = 4.00 \, \text{kg} + 3.00 \, \text{kg} + 2.00 \, \text{kg} = 9.00 \, \text{kg}

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