Chapter 7, Problem 49A

To determine

Force required to pull the toy boat on Jupiter.

Answer to Problem 49A

Force required to pull the toy boat on Jupiter is $F_{f\text{J}}=101.18\text{N}$ .

Explanation of Solution

Given:

Force required to pull toy boat on Earth is $F_f=40\text{N}$

Mass of toy boat is $m_b=10\text{kg}$

Formula used:

As,

  $\mu =\frac{F_f}{F_N}=\frac{F_f}{m_{b}g}$    …..     (1)

Here, $m_b$ is the mass of toy boat.

Normal force on Jupiter is given by:

  $F_N=G\frac{m_b{m}_J}{r_J{}^2}$ …..    (2)

As, $\mu =\frac{F_f}{F_N}$

So, $F_{fJ}=\mu F_N=\mu G\frac{m_b{m}_J}{r_J{}^2}$

Since,

  $\mu =\frac{F_f}{m_{b}g}$

So,

  $F_{fJ}=\frac{F_f}{m_{b}g}G\frac{m_b{m}_J}{r_J{}^2}=F_{f}G\frac{m_J}{g{r}_J{}^2}$

Calculation:

As,

  $\begin{array}{l}{F}_f=40\text{N}\\ G=6.67\times {10}^{-11}{\text{N-m}}^2/{\text{kg}}^{\text{2}}\\ m_J=1.9\times {10}^{27}\text{kg}\\ r_J=7.15\times {10}^7\text{m}\end{array}$

Putting the values

  $\begin{array}{l}{F}_{fJ}=40\times \left(6.67\times {10}^{-11}\right)\times \frac{1.9\times {10}^{27}}{9.8\times {\left(7.15\times {10}^7\right)}^2}\\ =101.18\text{N}\end{array}$

Conclusion:

Therefore, force required to pull the toy boat on Jupiter is $F_{f\text{J}}=101.18\text{N}$ .