Questions: 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. Two forces, F1 and F2, are applied to diagonally opposite comers, and act along the sides of the square, first as shown in case (i) and then as shown in case (ii) of the drawing. In each case the net torque produced by the forces is zero. If the magnitude of F2 is 6 times that of F1, find the distances a and b that locate the axis. It should be noted that a and b are not drawn to scale. a= Since we need to calculate torques, see if you can write an expression for the torque in terms of the force producing that torque and the perpendicular distance between the line of action of the force and the point about which we wish to determine the torque. After equating the clockwise to the counterclockwise torques for each case and inserting the relationship between the magnitude of the forces, we have two equations and two unknowns (σ and b). To determine the value for a, see if you can manipulate these two equations in such a manner as to eliminate b and allow you to solve for a. b= Similar to your calculation of a, after equating the clockwise to the counterclockwise torques for each case and inserting the relationship between the magnitude of the forces, we have two equations and two unknowns (α and b). To determine the value for b, see if you can manipulate these two equations in such a manner as to eliminate a and allow you to solve for b.

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. Two forces, F1 and F2, are applied to diagonally opposite comers, and act along the sides of the square, first as shown in case (i) and then as shown in case (ii) of the drawing. In each case the net torque produced by the forces is zero. If the magnitude of F2 is 6 times that of F1, find the distances a and b that locate the axis. It should be noted that a and b are not drawn to scale. a= Since we need to calculate torques, see if you can write an expression for the torque in terms of the force producing that torque and the perpendicular distance between the line of action of the force and the point about which we wish to determine the torque. After equating the clockwise to the counterclockwise torques for each case and inserting the relationship between the magnitude of the forces, we have two equations and two unknowns (σ and b). To determine the value for a, see if you can manipulate these two equations in such a manner as to eliminate b and allow you to solve for a. b= Similar to your calculation of a, after equating the clockwise to the counterclockwise torques for each case and inserting the relationship between the magnitude of the forces, we have two equations and two unknowns (α and b). To determine the value for b, see if you can manipulate these two equations in such a manner as to eliminate a and allow you to solve for b.
Transcript text: 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. Two forces, $\overrightarrow{\mathrm{F}}_{1}$ and $\overrightarrow{\mathrm{F}}_{2}$, are applied to diagonally opposite comers, and act along the sides of the square, first as shown in case (i) and then as shown in case (ii) of the drawing. In each case the net torque produced by the forces is zero. If the magnitude of $\vec{F}_{2}$ is 6 times that of $\vec{F}_{1}$, find the distances $a$ and $b$ that locate the axis. It should be noted that $a$ and $b$ are not drawn to scale. $a=$ $\square$ Since we need to calculate torques, see if you can write an expression for the torque in terms of the force producing that torque and the perpendicular distance between the line of action of the force and the point about which we wish to determine the torque. After equating the clockwise to the counterclockwise torques for each case and inserting the relationship between the magnitude of the forces, we have two equations and two unknowns ( $\sigma$ and $b$ ). To determine the value for $a$, see if you can manipulate these two equations in such a manner as to eliminate $b$ and allow you to solve for a. \[ b= \] Similar to your calculation of a, after equating the clockwise to the counterclockwise torques for each case and inserting the relationship between the magnitude of the forces, we have two equations and two unknowns ( $\alpha$ and $b$ ). To determine the value for $b$, see if you can manipulate these two equations in such a manner as to eliminate $a$ and allow you to solve for b.
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Solution

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Solution Steps

Step 1: Understanding the Problem

We have a square with side length 1 meter, and two forces \( \vec{F_1} \) and \( \vec{F_2} \) applied at diagonally opposite corners. The forces act along the sides of the square. The axis of rotation is perpendicular to the plane of the square and located at distances \( a \) and \( b \) from the sides. We need to find \( a \) and \( b \) such that the net torque is zero, given that \( F_2 \) is 6 times \( F_1 \).

Step 2: Setting Up Torque Equations

For the net torque to be zero, the sum of the torques produced by \( \vec{F_1} \) and \( \vec{F_2} \) must be zero. Torque (\( \tau \)) is given by \( \tau = r \times F \), where \( r \) is the perpendicular distance from the axis to the line of action of the force.

Step 3: Calculating Torques for Case (i)

In case (i), the torques produced by \( \vec{F_1} \) and \( \vec{F_2} \) are:

  • Torque by \( \vec{F_1} \): \( \tau_1 = a \cdot F_1 \)
  • Torque by \( \vec{F_2} \): \( \tau_2 = b \cdot F_2 \)

Given \( F_2 = 6F_1 \), we have: \[ \tau_2 = b \cdot 6F_1 \]

For equilibrium (net torque = 0): \[ a \cdot F_1 = b \cdot 6F_1 \] \[ a = 6b \]

Step 4: Calculating Torques for Case (ii)

In case (ii), the torques produced by \( \vec{F_1} \) and \( \vec{F_2} \) are:

  • Torque by \( \vec{F_1} \): \( \tau_1 = b \cdot F_1 \)
  • Torque by \( \vec{F_2} \): \( \tau_2 = a \cdot 6F_1 \)

For equilibrium (net torque = 0): \[ b \cdot F_1 = a \cdot 6F_1 \] \[ b = 6a \]

Step 5: Solving the System of Equations

From the two equations:

  1. \( a = 6b \)
  2. \( b = 6a \)

Substituting \( b = 6a \) into \( a = 6b \): \[ a = 6(6a) \] \[ a = 36a \]

This implies \( a = 0 \), which is not possible. Therefore, we need to re-evaluate the problem constraints or assumptions.

Final Answer

The problem constraints lead to a contradiction, suggesting a re-evaluation of the assumptions or additional information is needed.

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