Problem 81 - College Physics

Question

Two cups are placed on a uniform board that is balanced on a cylinder ( \(\mathbf{v}\) Fig. 6.40 ). The board has a mass of \(2.00 \mathrm{~kg}\) and is \(2.00 \mathrm{~m}\) long. The mass of \(\operatorname{cup} 1\) is \(200 \mathrm{~g}\) and it is placed \(1.05 \mathrm{~m}\) to the left of the balance point. The mass of cup 2 is \(400 \mathrm{~g}\). Where should cup 2 be placed for balance (relative to the right end of the board)?

Step-by-Step Solution

Answer

Cup 2 should be placed 0.475 m from the right end of the board.

Step 1: Convert Units

First, note that the mass of cup 1 is given in grams. To perform our calculations, we need all masses in kilograms. Convert the mass of cup 1 to kilograms: \[ 200 \text{ g} = 0.200 \text{ kg} \]. Similarly, convert the mass of cup 2: \[ 400 \text{ g} = 0.400 \text{ kg} \].

Step 2: Identify the Fulcrum and Balance Point

Since the board is uniform, its center of gravity is at the midpoint of the board (i.e., at 1.00 m from either end). Cup 1 is placed 1.05 m to the left of this point.

Step 3: Understand Torque for Static Equilibrium

For the board to be balanced, the net torque about the balance point (fulcrum) must be zero. This means the clockwise torque caused by cup 1 must be balanced by the counterclockwise torque caused by cup 2.

Step 4: Calculate Torque of Cup 1

Torque is calculated as the product of force and distance from the pivot (fulcrum). Torque by cup 1 is: \[ \tau_1 = m_1 \cdot g \cdot d_1 = 0.200 \cdot 9.8 \cdot 1.05 = 2.058 \text{ Nm} \], where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.

Step 5: Setup Equation for Torque Balancing

Let the distance of cup 2 from the fulcrum be \( x \) meters to the right. Then the counterclockwise torque by cup 2 is: \[ \tau_2 = m_2 \cdot g \cdot x = 0.400 \cdot 9.8 \cdot x = 3.92x \text{ Nm} \]. Equating the torques for balance: \[ 3.92x = 2.058 \].

Step 6: Solve for Distance \( x \)

Solve for \( x \) by dividing both sides of the equation by 3.92: \[ x = \frac{2.058}{3.92} \approx 0.525 \text{ m} \].

Step 7: Determine Location Relative to Board's Right End

Since the fulcrum is at 1.00 m from either end, and \( x = 0.525 \text{ m} \) from the fulcrum to the right, cup 2 should be placed: \[ 2.00 \text{ m} - (1.00 \text{ m} + 0.525 \text{ m}) = 0.475 \text{ m} \] from the right end of the board.

Key Concepts

Center of Gravity

The center of gravity is a crucial concept when studying the balance of objects. It is the point where the entire weight of an object appears to act. For a uniform board, like the one mentioned in the exercise, the center of gravity is located at its midpoint. In our scenario, the board is 2.00 meters long which places the center of gravity at exactly 1.00 meter from each end.
This midpoint is not dependent on the weight or the placement of the cups but rather on the uniformity and shape of the board itself. Understanding the position of the center of gravity allows us to determine the pivot point or fulcrum around which torques, or rotational forces, will act.
In problems involving static equilibrium and torque, identifying the center of gravity helps to simplify calculations as it indicates where we can assume the weight of the board is concentrated.

Static Equilibrium

Static equilibrium occurs when all forces and torques acting on a system result in no movement. For our seesaw-like system, this means it remains perfectly balanced. To achieve this balance, the sum of the clockwise torques must equal the sum of the counterclockwise torques.
Imagine you are on a seesaw. If both sides weigh equally and are evenly distanced from the pivot, the seesaw remains level. This is a basic example of static equilibrium. In physics problems, achieving static equilibrium involves careful attention to the magnitudes and directions of forces, and their distances from the fulcrum.
In our exercise, the board and cups reach static equilibrium when the torques produced by the weights of the cups on either side of the fulcrum are equal, preventing the board from tipping over.

Torque Calculation

Torque is the turning effect of a force applied at a distance from a pivot point, and is crucial in achieving balance in the exercise. It is calculated using the formula: \[ \tau = m \cdot g \cdot d \] where \( \tau \) is the torque, \( m \) is the mass, \( g \) is the gravitational acceleration (approximately 9.8 m/s² on Earth), and \( d \) is the perpendicular distance from the pivot point.
When calculating torque, remember that it depends not only on the amount of force but also on how far from the fulcrum this force is applied. This means that even a small mass can exert a large torque if it's placed far enough away from the balance point.
As seen in the exercise, cup 1 creates a torque that needs to be countered by the torque from cup 2. This involves manipulating the equations to find the perfect placement of cup 2 to ensure the board remains in static equilibrium, helping explain where exactly to place cup 2 for achieving balance.