Questions: 3. Two cats are on a seesaw, which is a 5.5 -meter long wood plank. The orange cat has a mass of 5.93 kg and is 0.361 meters from the edge of the plank. The black cat is sitting at 2.71 meters from the pivot point, what is the weight of the black cat?

Solution Steps

The problem involves a seesaw in equilibrium with two cats on it. We need to find the weight of the black cat given the position and weight of the orange cat.

• Length of the plank: 5.5 meters
• Mass of the orange cat: 5.93 kg
• Distance of the orange cat from the edge: 0.361 meters
• Distance of the black cat from the pivot: 2.71 meters

Assume the pivot point is at the center of the plank, which is at 2.75 meters from either edge (since \( \frac{5.5}{2} = 2.75 \)).

The distance of the orange cat from the pivot is \( 2.75 - 0.361 = 2.389 \) meters.

For the seesaw to be in equilibrium, the moments about the pivot must be equal. The moment is calculated as the product of force (weight) and distance from the pivot.

The moment due to the orange cat is \( 5.93 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 2.389 \, \text{m} \).

The moment due to the black cat is \( m \times 9.81 \, \text{m/s}^2 \times 2.71 \, \text{m} \), where \( m \) is the mass of the black cat.

Set the moments equal to each other: \[ 5.93 \times 9.81 \times 2.389 = m \times 9.81 \times 2.71 \]

Cancel out \( 9.81 \) from both sides and solve for \( m \): \[ 5.93 \times 2.389 = m \times 2.71 \]

\[ m = \frac{5.93 \times 2.389}{2.71} \]

Compute the value of \( m \) to find the mass of the black cat.

Final Answer