Chapter 7, Problem 49P

To determine

The CM of the loaded ferryboat.

Answer to Problem 49P

The CM of the loaded ferryboat is $\left(3\text{ m, }-3\text{ m}\right)$ .

Explanation of Solution

Given:

The size of raft is $18\text{ m}\times 18\text{ m}$ .

The mass of ferry boat is $M=6,800\text{ kg}$ .

The mass of the three cars is $m_1=m_2=m_3=1,200\text{ kg}$ .

Formula used:

The expression for the distance of CM $\left(x_{CM}\right)$ is,

  $x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

Here, $m_1$ is the mass of the object 1, $m_2$ is the mass of object 2, $m_3$ is the mass of object 3, $x_1$ is the distance of object 1 from the origin along x direction, $x_2$ is the distance of object 2 from the origin along x direction, and $x_3$ is the distance of object 3 from the origin along x direction.

The expression for the distance of CM $\left(y_{CM}\right)$ is,

  $y_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$

Here, $m_1$ is the mass of the object 1, $m_2$ is the mass of object 2, $m_3$ is the mass of object 3, $y_1$ is the distance of object 1 from the origin along y direction, $y_2$ is the distance of object 2 from the origin along y direction, and $y_3$ is the distance of object 3 from the origin along y direction.

Calculation:

Sketch the representation as shown below.

  

The position of mass 1 is $\left(x_1,y_1\right)=\left(9\text{ m, 9 m}\right)$ .

The position of mass 2 is $\left(x_2,y_2\right)=\left(9\text{ m, }-\text{9 m}\right)$ .

The position of mass 2 is $\left(x_3,y_3\right)=\left(-9\text{ m, }-\text{9 m}\right)$ .

The distance of CM along x axis is,

  $\begin{array}{l}{x}_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2}\\ x_{\text{CM}}=\frac{\left(\text{1,200 kg}\right)\left(\text{9 m}\right)+\left(\text{1,200 kg}\right)\left(9\text{ m}\right)+\left(\text{1,200 kg}\right)\left(-9\text{ m}\right)}{\left(\text{1,200 kg}\right)+\left(\text{1,200 kg}\right)+\left(\text{1,200 kg}\right)}\\ x_{\text{CM}}=3\text{ m}\end{array}$

The distance of CM along y axis is,

  $\begin{array}{l}{y}_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2}\\ y_{\text{CM}}=\frac{\left(\text{1,200 kg}\right)\left(\text{9 m}\right)+\left(\text{1,200 kg}\right)\left(-9\text{ m}\right)+\left(\text{1,200 kg}\right)\left(-9\text{ m}\right)}{\left(\text{1,200 kg}\right)+\left(\text{1,200 kg}\right)+\left(\text{1,200 kg}\right)}\\ y_{\text{CM}}=-3\text{ m}\end{array}$

Conclusion:

Thus, the CM of the loaded ferryboat is $\left(x_{\text{CM}},y_{\text{CM}}\right)=\left(3\text{ m, }-3\text{ m}\right)$ .