### Rotational Dynamics Concept – Torque and EquilibriumAs shown in the figure below, we have a square one meter on a side that is free to rotate about an axis perpendicular to the plane of the square, a distance \( a \) from one side and a distance \( b \) from the other side.**Diagrams:**1. **Figure (a)** - Square with forces \( \vec{F}_1 \) and \( \vec{F}_2 \) applied at diagonally opposite corners. - \( \vec{F}_1 \) is directed upwards on the left corner. - \( \vec{F}_2 \) is directed downwards on the right corner.2. **Figure (b)** - Square with forces \( \vec{F}_1' \) and \( \vec{F}_2' \) applied at diagonally opposite corners. - \( \vec{F}_1' \) is directed upwards on the left corner. - \( \vec{F}_2' \) is directed to the right on the right corner.Both diagrams involve a square viewed in a two-dimensional plane with the forces clearly marked, indicating the direction of forces along the sides of the square. These forces contribute to the rotational torque about the specified axis.### Mathematical Explanation:Two forces, \( \vec{F}_1 \) and \( \vec{F}_2 \), are applied to diagonally opposite corners, and act along the sides of the square, first as shown in case (a) and then as shown in case (b) of the drawing. In each case, the net torque produced by the forces is zero. If the magnitude of \( \vec{F}_1 \) is 6 times that of \( \vec{F}_2 \), find the distances \( a \) and \( b \) that locate the axis. It should be noted that \( a \) and \( b \) are not drawn to scale.#### Calculations:- Let \( \vec{F}_2 = F \)- Then \( \vec{F}_1 = 6F \)**Determine Distances \( a \) and \( b \):**\[ a = \_ \_ \_ \_ \_ \] m\[ b = \_ \_ \_ \_ \_ \] mBy solving the respective torque equilibrium

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As shown in the figure below, we have a square one meter on a side that is free to rotate about an axis perpendicular to the plane of the square, a distance a from one side and a distance b from the other side. (1) Two forces, F, and F are applied to diagonally opposite coners, and act along the sides of the square, first as shown in case (1) and then as shown in case () of the drawing. In each case the net torque produced by the forces is zero. If the magnitude of F, is s times that of E, find the distances a and b that locate the axis. It should be noted that a and b are not drawn to scale. m m

As shown in the figure below, we have a square one meter on a side that is free to rotate about an axis perpendicular to the plane of the square, a distance a from one side and a distance b from the other side. (1) Two forces, F, and F are applied to diagonally opposite coners, and act along the sides of the square, first as shown in case (1) and then as shown in case () of the drawing. In each case the net torque produced by the forces is zero. If the magnitude of F, is s times that of E, find the distances a and b that locate the axis. It should be noted that a and b are not drawn to scale. m m

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

Rigid Body

A rigid body is an object which does not change its shape or undergo any significant deformation due to an external force or movement. Mathematically speaking, the distance between any two points inside the body doesn't change in any situation.

Rigid Body Dynamics

Rigid bodies are defined as inelastic shapes with negligible deformation, giving them an unchanging center of mass. It is also generally assumed that the mass of a rigid body is uniformly distributed. This property of rigid bodies comes in handy when we deal with concepts like momentum, angular momentum, force and torque. The study of these properties – viz., force, torque, momentum, and angular momentum – of a rigid body, is collectively known as rigid body dynamics (RBD).

Question

### Rotational Dynamics Concept – Torque and Equilibrium

As shown in the figure below, we have a square one meter on a side that is free to rotate about an axis perpendicular to the plane of the square, a distance \( a \) from one side and a distance \( b \) from the other side.

**Diagrams:**
1. **Figure (a)**
- Square with forces \( \vec{F}_1 \) and \( \vec{F}_2 \) applied at diagonally opposite corners.
- \( \vec{F}_1 \) is directed upwards on the left corner.
- \( \vec{F}_2 \) is directed downwards on the right corner.

2. **Figure (b)**
- Square with forces \( \vec{F}_1' \) and \( \vec{F}_2' \) applied at diagonally opposite corners.
- \( \vec{F}_1' \) is directed upwards on the left corner.
- \( \vec{F}_2' \) is directed to the right on the right corner.

Both diagrams involve a square viewed in a two-dimensional plane with the forces clearly marked, indicating the direction of forces along the sides of the square. These forces contribute to the rotational torque about the specified axis.

### Mathematical Explanation:
Two forces, \( \vec{F}_1 \) and \( \vec{F}_2 \), are applied to diagonally opposite corners, and act along the sides of the square, first as shown in case (a) and then as shown in case (b) of the drawing. In each case, the net torque produced by the forces is zero. If the magnitude of \( \vec{F}_1 \) is 6 times that of \( \vec{F}_2 \), find the distances \( a \) and \( b \) that locate the axis. It should be noted that \( a \) and \( b \) are not drawn to scale.

#### Calculations:
- Let \( \vec{F}_2 = F \)
- Then \( \vec{F}_1 = 6F \)

**Determine Distances \( a \) and \( b \):**

\[ a = \_ \_ \_ \_ \_ \] m
\[ b = \_ \_ \_ \_ \_ \] m

By solving the respective torque equilibrium

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