You have just planted a sturdy \({\rm{2}}\;{\rm{m}}\)tall palm tree in your front lawn for your mother’s birthday. Your brother kicks a \({\rm{500}}\;{\rm{g}}\) ball, which hits the top of the tree at a speed of \({\rm{5}}\;{\rm{m/s}}\)and stays in contact with it for \({\rm{10}}\;{\rm{ms}}\). The ball falls to the ground near the base of the tree and the recoil of the tree is minimal. (a) What is the force on the tree? (b) The length of the sturdy section of the root is only \({\rm{20}}\;{\rm{cm}}\). Furthermore, the soil around the roots is loose and we can assume that an effective force is applied at the tip of the \({\rm{20}}\;{\rm{cm}}\) length. What is the effective force exerted by the end of the tip of the root to keep the tree from toppling? Assume the tree will be uprooted rather than bend. (c) What could you have done to ensure that the tree does not uproot easily?

In equilibrium, the total clockwise torque equals the total anti-clockwise torque.

The height of the palm tree is \({{\rm{r}}_{\rm{1}}}{\rm{ = 2}}\;{\rm{m}}\).

The mass of the ball is \({\rm{m = 500}}\;{\rm{g}}\).

The speed of the ball is \({{\rm{v}}_{\rm{1}}}{\rm{ = 5}}\;{\rm{m/s}}\).

The time of contact is \({\rm{t = 10}}\;{\rm{ms}}\).

The length of the root is \({{\rm{r}}_{\rm{2}}}{\rm{ = 20}}\;{\rm{cm}}\).

\(\begin{align}{\rm{\Delta p = 0 - m}}{{\rm{v}}_{\rm{1}}}\\{\rm{\Delta p = - }}\frac{{{\rm{500}}}}{{{\rm{1000}}}}{\rm{ \times 5}}\\{\rm{\Delta p = - 2}}{\rm{.50}}\;{\rm{kg \times m/s}}\end{align}\)

The force by the palm tree on the ball is,

\(\begin{align}{{\rm{F}}_{\rm{r}}}{\rm{ = }}\frac{{{\rm{\Delta p}}}}{{\rm{t}}}\\{\rm{ = - }}\frac{{{\rm{2}}{\rm{.50}}}}{{{\rm{10 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}}}\\{\rm{ = - 250}}\;{\rm{N}}\end{align}\)

Hence, the force on the tree is the same but opposite by the force on the ball which is,

\({\rm{F = 250}}\;{\rm{N}}\).

\(\begin{align}{\rm{F \times }}{{\rm{r}}_{\rm{1}}}{\rm{ + }}{{\rm{F}}_{\rm{e}}}{\rm{ \times }}{{\rm{r}}_{\rm{2}}}{\rm{ = 0}}\\{\rm{250 \times 2 - }}{{\rm{F}}_{\rm{e}}}{\rm{ \times }}\frac{{{\rm{20}}}}{{{\rm{100}}}}{\rm{ = 0}}\\{{\rm{F}}_{\rm{e}}}{\rm{ = }}\frac{{{\rm{250 \times 2}}}}{{\frac{{{\rm{20}}}}{{{\rm{100}}}}}}\\{{\rm{F}}_{\rm{e}}}{\rm{ = 2500}}\;{\rm{N}}\end{align}\)

Hence, the force is \({\rm{2500}}\;{\rm{N}}\).

c. Compressing the soil at the tree's base provides an extra reaction force, eases the uprooting.