Analyzing Forces on a Uniform Meter Scale in Equilibrium

This problem involves a uniform meter scale supported by two wedges, which is a classic example of a rigid body in static equilibrium. For the meter scale to be in equilibrium, two main conditions must be satisfied:

1. The net force acting on the scale must be zero.
2. The net torque acting on the scale about any point must be zero.

Problem Setup and Identifying Forces

We have a uniform meter scale of mass m = 0.24 kg. Since it is uniform, its center of mass (CM) is located at the 0.5 m mark from either end. The weight of the scale acts downwards at the CM.

• Mass of the scale, \(m = 0.24\) kg
• Acceleration due to gravity, \(g = 10.0\) m s^-2
• Weight of the scale, \(W = mg = 0.24 \, \text{kg} \times 10.0 \, \text{m s}^{-2} = 2.4\) N. This acts downwards at the 0.5 m mark.

The scale is supported by two wedges, W₁ and W₂, which exert upward normal forces N₁ and N₂ respectively.

• W₁ is at a distance of 0.2 m from one end. Let's assume this is the left end (0 m mark). So, W₁ is at the 0.2 m mark. The upward force is N₁.
• W₂ is at a distance of 0.4 m from the other end. The other end is the right end (1 m mark). So, W₂ is at \(1.0 \, \text{m} - 0.4 \, \text{m} = 0.6\) m mark. The upward force is N₂.

The forces acting on the scale are:

• Weight \(W = 2.4\) N downwards at the 0.5 m mark.
• Normal force \(N_1\) upwards at the 0.2 m mark.
• Normal force \(N_2\) upwards at the 0.6 m mark.

Applying Conditions for Static Equilibrium

1. Translational Equilibrium (Net Force is Zero)

The sum of the upward forces must equal the sum of the downward forces. \[N_1 + N_2 = W\] \[N_1 + N_2 = 2.4 \, \text{N} \quad \text{(Equation 1)}\]

2. Rotational Equilibrium (Net Torque is Zero)

The sum of the torques about any point must be zero. Choosing a convenient pivot point can simplify calculations. Let's choose the position of W₁ (at 0.2 m) as the pivot point.

Torque (\(\tau\)) is calculated as force multiplied by the perpendicular distance from the pivot to the line of action of the force.

• Torque due to N₁: The force \(N_1\) acts at the pivot point, so the distance is zero. Torque \(\tau_1 = N_1 \times 0 = 0\).
• Torque due to W: The weight \(W\) acts at the 0.5 m mark. The distance from the pivot (0.2 m mark) is \(0.5 \, \text{m} - 0.2 \, \text{m} = 0.3\) m. This force tends to cause a clockwise rotation about the pivot. Torque \(\tau_W = W \times 0.3 \, \text{m}\).
• Torque due to N₂: The force \(N_2\) acts at the 0.6 m mark. The distance from the pivot (0.2 m mark) is \(0.6 \, \text{m} - 0.2 \, \text{m} = 0.4\) m. This force tends to cause an anti-clockwise rotation about the pivot. Torque \(\tau_2 = N_2 \times 0.4 \, \text{m}\).

For rotational equilibrium, the sum of clockwise torques equals the sum of anti-clockwise torques.

Substitute the value of W:

Now, solve for \(N_2\):

Calculating N₁

Now that we have the value for \(N_2\), we can use Equation 1 (from translational equilibrium) to find \(N_1\):

Summary of Results

The forces exerted by the wedges are \(N_1 = 0.6\) N and \(N_2 = 1.8\) N.

This calculation shows that the force N₂ is greater than N₁, which makes sense as the weight's center of mass is closer to W₂ (0.5m is 0.1m away from 0.6m) than it is to W₁ (0.5m is 0.3m away from 0.2m).

Revision Table: Key Concepts

Additional Information: Choosing a Pivot Point

In a static equilibrium problem involving torque, you can choose any point as the pivot. The net torque about any point must be zero if the object is in rotational equilibrium. Choosing one of the points where an unknown force acts (like the location of a wedge in this case) is often strategic because the torque due to that force about that pivot is zero, simplifying the torque equation. Let's quickly check if we chose W₂ (0.6 m) as the pivot:

• Torque due to N₂: 0 (at the pivot).
• Torque due to W: W is at 0.5m. Distance from pivot (0.6m) is \(0.6 \, \text{m} - 0.5 \, \text{m} = 0.1\) m. Torque \(= W \times 0.1\) (Anti-clockwise).
• Torque due to N₁: N₁ is at 0.2m. Distance from pivot (0.6m) is \(0.6 \, \text{m} - 0.2 \, \text{m} = 0.4\) m. Torque \(= N_1 \times 0.4\) (Clockwise).

Equating torques: \(N_1 \times 0.4 \, \text{m} = W \times 0.1 \, \text{m}\)

\(N_1 \times 0.4 \, \text{m} = 2.4 \, \text{N} \times 0.1 \, \text{m}\)

\(N_1 \times 0.4 \, \text{m} = 0.24 \, \text{Nm}\)

\(N_1 = \frac{0.24 \, \text{Nm}}{0.4 \, \text{m}} = 0.6 \, \text{N}\)

This gives the same value for N₁. Then using \(N_1 + N_2 = 2.4\), we get \(0.6 + N_2 = 2.4\), which gives \(N_2 = 1.8\) N. The results are consistent, confirming the validity of choosing any pivot point.