## Steel Cables

Consider a 200 ft long steel cable is used to help hikers climb up Half Dome in Yosemite National Park, CA. During the summer, the ambient temperature can reach up to 95°F. If the cable shrinks by a max of 1.2 inches in the winter due to thermal deformation, how cold does Half Dome get in °F? Note the coefficient of thermal expansion of steel is 0.000007 ft/ft°F.

##
__
__**Hint**

**Hint**

$$$\delta_t=\alpha L(T-T_0)$$$

where
$$\delta_t$$
is the deformation caused by a temp change,
$$\alpha$$
is the temp coefficient of expansion,
$$L$$
is the member length,
$$T$$
is the final temp, and
$$T_0$$
is the initial temp.

##
__
__**Hint 2**

**Hint 2**

Solve for
$$T$$
:

$$$T=\frac{\delta_t}{\alpha L}+T_0$$$

The equation for thermal deformation is:

$$$\delta_t=\alpha L(T-T_0)$$$

where
$$\delta_t$$
is the deformation caused by a temp change,
$$\alpha$$
is the temp coefficient of expansion,
$$L$$
is the member length,
$$T$$
is the final temp, and
$$T_0$$
is the initial temp.

Solving for
$$T$$
:

$$$T=\frac{\delta_t}{\alpha L}+T_0$$$

Remember,
$$\delta_t$$
is negative because the cables shrink in the winter and to convert the 1.2 inch shrinkage to feet. Thus,

$$$T=\frac{-0.1ft(ft^{\circ}F)}{0.000007ft (200ft)}+95^{\circ}F=\frac{-14,285.74^{\circ}F}{200}+95^{\circ}F$$$

$$$=-71.43^{\circ}F+95^{\circ}F=23.6^{\circ}F$$$

23.6°F