Question

A 0.120-kg, 50.0-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.110-kg mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

Short answer

The fulcrum should be placed approximately 29.82 cm from the left end.

Step-by-step solution

Determine the Position of the Center of the Uniform Bar

The center of gravity (CG) for a uniform bar is located at its geometric center. Since the bar is 50.0 cm long, its center will be at 25.0 cm from either end. Let us denote this position as \( x_b = 25.0 \text{ cm} \).

Calculate the Moments about a Chosen Point

We will calculate the moments about the left end of the bar, which we will call point O. We need to consider the uniform bar and the two point masses at the ends. The formula for the moment of a mass \( m \) located at position \( x \) is given by \( m \cdot x \). Thus:1. For the 0.055-kg mass at the left: \( M_1 = 0.055 \times 0 = 0 \).2. For the 0.110-kg mass on the right: \( M_2 = 0.110 \times 50.0 = 5.5 \text{ kg} \cdot \text{cm} \).3. For the 0.120-kg bar: \( M_b = 0.120 \times 25.0 = 3.0 \text{ kg} \cdot \text{cm} \).

Calculate the Total Moment about Point O

To find the center of gravity, we need the balance of moments to equal zero, meaning the sum of all moments must be used:\[ \text{Total Moment} = M_1 + M_b + M_2 \]\[ = 0 + 3.0 + 5.5 = 8.5 \text{ kg} \cdot \text{cm} \].

Calculate the Total Mass of the System

Add up the masses of all components:1. Mass of 0.055 kg glue: \( m_1 = 0.055 \text{ kg} \).2. Mass of the bar: \( m_b = 0.120 \text{ kg} \).3. Mass of 0.110 kg glue: \( m_2 = 0.110 \text{ kg} \).\( \text{Total Mass} = m_1 + m_b + m_2 = 0.055 + 0.120 + 0.110 = 0.285 \text{ kg} \).

Calculate the Center of Gravity from the Left End

The center of gravity position \( x_{cg} \) can be calculated by dividing the total moment by the total mass:\[ x_{cg} = \frac{\text{Total Moment}}{\text{Total Mass}} = \frac{8.5}{0.285} \approx 29.82 \text{ cm} \].

Key concepts

Uniform Bar

A uniform bar is a straight object that has the same cross-sectional dimensions and material properties along its entire length. This means that the mass is distributed evenly across the bar. In the context of physics problems, a uniform bar is often used to simplify calculations.
The uniform distribution of mass allows us to pinpoint the center of gravity (CG) easily. We calculate it to be precisely at the geometric center of the bar. For example, if a bar is 50.0 cm long, its CG will be at 25.0 cm from either end.
This concept helps us to understand how the weight of the bar is balanced. Such an understanding is crucial when performing further calculations like finding moments and balancing the bar on a fulcrum.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to rotation around an axis. It depends on how the object's mass is distributed with respect to the pivot point.
For a point mass, the moment of inertia can be calculated simply by multiplying the mass by the square of the distance from the pivot:

• For a point mass, the formula is given as: \[ I = m imes r^2 \]where \( I \) is the moment of inertia, \( m \) is the mass, and \( r \) is the distance from the distance to the axis of rotation.

When dealing with multiple masses, like a bar with point masses on either end, the moment of inertia for each mass must be considered separately.
By understanding the moment of inertia, we can calculate the total effects of all masses on the system's rotation.

Point Masses

Point masses refer to small objects whose size can be ignored, and they are considered as having their entire mass concentrated at a single point. This concept is useful in simplifying various calculations like those involving the center of gravity or moments.
In a physics problem, each point mass affects the balance of the system. We calculate the contribution of each point mass using their masses and their distances from a reference point. For example:

• A 0.055-kg point mass on the left contributes minimally since it rests very close to the reference point.
• A 0.110-kg mass on the right contributes significantly more as it affects the balance further away from the pivot.

Simplifying a system with point masses allows for straightforward calculations on how they will affect the whole system.

Fulcrum Balance

Fulcrum balance involves positioning the pivot point such that the system remains in equilibrium. This concept is essential when you want to balance asymmetrical systems, such as a bar with different masses glued to its ends.
The balance point, or the fulcrum, is typically placed directly beneath the center of gravity. In achieving balance, the sum of clockwise moments must equal the sum of counterclockwise moments around the fulcrum.

• Moments are calculated by multiplying the weight of each mass with its distance from the fulcrum.
• Adjusting the fulcrum position can re-distribute these moments to maintain equilibrium.

With the correct understanding of fulcrum balance, we can predictably determine where the fulcrum should be placed so that the entire system stays stable.