Chapter 9: Q16P (page 248)

Ricardo, of mass 80 kg , and Carmelita, who is lighter, is enjoying Lake Merced at dusk in a30 kgcanoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 mapart and symmetrically located with respect to the canoe’s center. If the canoe moves 40 cmhorizontally relative to a pier post, what is Carmelita’s mass?

Short Answer

Expert verified

The mass of the Carmelita is $m_{2} = 58 k g$ .

Step by step solution

Listing the given quantities:

The mass of Ricardo is $m_{1} = 80 k g$ .

The mass of the canoe is $m_{3} = 30 k g$ .

The separation distance between Ricardo and Carmelita is $L = 3.0 m$ .

The canoe shifted by the distance is $2 x = 0.40 m$ .

Understanding the concept of center of mass:

The center of gravity is a position defined relative to an object or system of objects. It is the average position of all parts of the system, weighted by their weight. For simple rigid objects of uniform density, the center of mass is located at the center of gravity.

You can use the concept of the center of mass of the system.

Formula:

$\vec{R_{c m}} = \frac{m_{1} \vec{r_{1}} + m_{2} \vec{r_{2}}}{m_{1} + m_{2}}$

Here, $\vec{R_{c m}}$ is the distance to the center of mass, ${\vec{r}}_{1}$ is the distance of mass $m_{1}$ , and ${\vec{r}}_{2}$ is the distance of mass $m_{2}$ .

Calculations for the mass of Carmelita:

The acceleration of the center of mass of the cart-block system:

You can apply the center of mass of the system concept.

Consider the mass of the Carmelita as .

$\vec{R_{c m}} = \frac{m_{1} \vec{r_{1}} + m_{2} \vec{r_{2}} + m_{3} \vec{r_{3}}}{m_{1} + m_{2} + m_{3}}$

The center of the mass is not moving, hence

role="math" localid="1661238681972" $0 = \frac{m_{1} (\frac{L}{2} - x) - m_{2} (\frac{L}{2} - x) - m_{3} x}{m_{1} + m_{2} + m_{3}} m_{3} x = m_{1} (\frac{L}{2} - x) - m_{2} (\frac{L}{2} - x) . . . . . . . . (1)$

When they change their positions, the center of the canoe is shifted by 2x from its initial position.

$2 x = 0.40 m x = 0.20 m$

Equation (1) becomes,

$m_{2} (\frac{L}{2} - x) = m_{1} (\frac{L}{2} - x) - m_{3} x m_{2} = \frac{m_{1} (\frac{L}{2} - x) - m_{3} x}{m_{2} (\frac{L}{2} - x)}$

Substitute known values in the above equation.

$m_{2} = \frac{80 k g (\frac{3.0 m}{2} - 0.20 m) - 30 k g \times 0.20 m}{(\frac{3.0 m}{2} + 0.20 m)} = \frac{104 k g . m - 6 k g . m}{1.7 m} = \frac{98 k g . m}{1.7 m} = 58 k g$