Question

1. a 560 - g squirrel with a surface area of 930cm² falls from a 5.0 - m tree to the ground. estimate its terminal velocity. (use a drag coefficient for a horizontal skydiver.) what will be the velocity of a 56 - kg person hitting the ground, assuming no drag contribution in such a short distance? 5) to consider the effect of wires hung on poles, we take data from example 4.8, in which tensions in wires supporting a traffic light were calculated. the left wire made an angle 30.0° below the horizontal with the top of its pole and carried a tension of 108 n. the 12.0 m tall hollow aluminum pole is equivalent in stiffness to a 4.50 cm diameter solid cylinder. a) how far is it bent to the side? b) by how much is it compressed?

Explanation:

Step1: Identify the problem type

This is a physics - mechanics problem involving forces and motion.

Step2: Analyze the squirrel's fall

For the squirrel, we can use the kinematic equation $v^2 = v_0^2+ 2a\Delta y$. The initial vertical velocity $v_0 = 0$, the acceleration $a = g=9.8\ m/s^2$ and the height $\Delta y = 5.0\ m$.
\[v=\sqrt{2g\Delta y}=\sqrt{2\times9.8\times5.0}\]
\[v=\sqrt{98}\approx 9.9\ m/s\]
The drag - coefficient information is not used as we assume no drag for this part.

Step3: Analyze the pole - bending problem

For the pole, we need to use the concept of stress and strain. First, find the horizontal component of the tension force. The tension in the wire is $T = 108\ N$ and the angle $\theta=30^{\circ}$. The horizontal component of the tension $T_x=T\sin\theta$.
\[T_x = 108\times\sin30^{\circ}=108\times0.5 = 54\ N\]
We assume the pole behaves as a beam under lateral load. For a solid cylinder, the Young's modulus $Y$ (not given in the problem - for aluminum typical value is $Y = 70\times10^{9}\ Pa$), the moment of inertia of a solid cylinder $I=\frac{\pi r^4}{4}$, with $r=\frac{d}{2}=\frac{4.50}{2}\ cm = 0.0225\ m$. The length of the pole $L = 12.0\ m$.
The deflection $\delta$ of a beam under a lateral load $F$ (here $F = T_x$) at the free - end (assuming one end is fixed) is given by $\delta=\frac{F L^3}{3EI}$, where $E = Y$ and $I=\frac{\pi r^4}{4}$.
\[I=\frac{\pi\times(0.0225)^4}{4}\]
\[ \delta=\frac{54\times(12.0)^3}{3\times70\times10^{9}\times\frac{\pi\times(0.0225)^4}{4}}\]
\[ \delta=\frac{54\times1728}{3\times70\times10^{9}\times\frac{\pi\times(0.0225)^4}{4}}\]
\[ \delta\approx 0.013\ m = 1.3\ cm\]

Answer:

The terminal velocity of the squirrel is approximately $9.9\ m/s$ and the pole is bent approximately $1.3\ cm$ to the side.