Introduction:

Please get ready to welcome the winter, with its snow and cold temps, is just around the corner. According to the National Oceanic and Atmospheric Administration (NOAA), New England can expect a milder season than normal. As humans we must have the ability to withstand and fight in against of all kinds of natural calamity, epidemic, pandemic, poverty, superstition, illiteracy, diseases, social and religious discrimination with our knowledge, technological skills, creativity and wisdom. We have to take care our health and mind with our own effort. Health care is through self regulation, self visualization, discipline, exercise, intaking nutritious and healthy food, maintaining required sleep, abstaining from alcohol, drugs, smoking, engaging in social charity and welfare activities, involve in education & research, awareness about social and political activities in and around the world. A mind with moral integrity is key to all success leading to eternity.

The Old Farmer’s Almanac based in New Hampshire, founded in 1792, has released its official U.S. Winter Forecast for 2020-2021.

“It’s Snow Time!” the almanac said. “Get excited, folks in the Northeast! We can’t spill the beans on the entire forecast, but snow lovers should be very excited!”

“Remember last year’s almost snow-free winter in the Northeast? Well, this year our prediction is very different, with the possibility of a blizzard hitting the Mid-Atlantic and Northeast states during the second week of February.”

The almanac said the storm could bring between 1 and 2 feet of snow to cities from Washington, D.C. up to Boston.

The forecast also predicts a big storm during the final week of March from the nation’s midsection to central New England, bringing “a significant late-season snowfall to the north of its track.”

Frictional Force:

Frictional forces help us to walk, car to move and wheel to rotate. Imagine our world is covered with ice or some kind of oily substance on which our normal movements couldn’t be maintained. Although a substantial percent (about 15-20%) of gasoline is used by an automobile or train to overcome the friction but without having friction we couldn’t get around with our moving devices, we couldn’t walk or ride a bicycle. We are not able hold things with our hands.

An object shows the following properties as it slides from rest to movement upon the application of a force:

Property 1. If the object does not move, then the static frictional force \(\vec f_s\) and the applied force \(\vec F\) are parallel to each other with equal magnitude and directed opposite to each other.

Property 2. The magnitude of \(\vec f_s\) has a maximum value \(f{s,max}\) that is given by

\[\begin{equation} \tag{1} f_{s,max} = \mu_sF_N \end{equation}\]

where \(\mu_s\) is the coefficient of static friction and \(F_N\) is the magnitude of the normal force on the body from the surface. If the magnitude of the component of \(\vec F\) that is parallel to the surface exceeds \(f_{s,max}\), then the body begins to slide along the surface.

Property 3. Once the object begins to slide along the surface, the magnitude of the frictional force rapidly decreases to a value \(f_k\) and is given by

\[\begin{equation} \tag{2} f_k = \mu_kF_N \end{equation}\]

where \(\mu_k\) is the coefficient of kinetic friction.

Example Problem 6.1: Stopping Distances for a Sliding Car

Calculate the stopping distances for a car sliding to a stop from an initial speed of 10.0 m/s (36 km/h) on a dry horizontal road, an icy horizontal road, and an icy downhill.

  1. How far does the car slides to a stop on a horizontal road (Fig. 6-1a) if the coefficient of kinetic friction is \(\mu_k = 0.60\)? Assume that the air resistance is negligibly small on the car, and also assume that the wheels of the car are locked up and the wheels slide, along the x-direction of the car’s movement.

  2. What is the stopping distance if the road is covered with ice with \(\mu_k = 0.10\)?

  3. Repeat the calculation assuming that the car sliding down on an icy hill with an angle of inclination \(\theta = 5.0^\circ.\)

Solution:

  1. Using free-body diagram of Figure 6-1b, we get

\[\begin{equation} \tag{3} -f_k=ma_x \end{equation}\]

Substituting \(f_k = \mu_kF_N\) and \(F_N = mg\) give us

\[\begin{equation} \tag{4} -\mu_kF_N=ma_x \end{equation}\]

\[\begin{equation} \tag{5} a_x= -\mu_kg. \end{equation}\]

Using Equation, \(v^2 = v_0^2+2a(x-x_0)\), we get \(x-x_0=\frac{(v^2-v_0^2)}{2a_x}\)

\[\begin{equation} \tag{6} x-x_0 = \frac{(v^2-v_0^2)}{-2\mu_kg} \end{equation}\]

Now substituting v = 0, \(v_0 = 10~m/s\), \(\mu_k=0.60\), and \(g=9.8~m/s^2\), we get \(x-x_0=\frac{(v^2-v_0^2)}{-2\mu_kg}=\frac{(0-(10~m/s)^2)}{-2\times 0.6\times 9.8~m/s}=8.5~m\space(Answer)\)

  1. Changing the value of \(\mu_k = 0.1\) for ice, we get \(x-x_0=\frac{(v^2-v_0^2)}{-2\mu_kg}=\frac{(0-(10~m/s)^2)}{-2\times 0.1\times 9.8~m/s}=51~m.\space(Answer)\)

  2. Resolving the forces along the x and y-components of Figure 6-1c, we get \[-f_k+mg\sin\theta=ma_x\]

\[-\mu_kF_N+mg\sin\theta=ma_x\]

\[F_N=mg\cos\theta\]

\[-mu_k(mg\cos\theta)+mg\sin\theta=ma_x\]

\[a_x=-\mu_kg\cos\theta+g\sin\theta=-(0.10)(9.8~m/s^2)\cos5^\circ+(9.8~m/s^2)\sin5^\circ)=-0.122~m/s^2\]

\[x-x_0=\frac{v^2-v_0^2)}{2a_x}=\frac{(0-(10~m/s)^2)}{2(-0.122~m/s^2)}=400~m\space(Answer)\]

Caution!!. This result is a cautionary signal for the drivers. Be careful of driving on the icy roads and highways during winter. In this example the stopping distances are 8.5 m, 51 m, and 400 m for a car sliding to a stop from an initial speed of 10.0 m/s (36 km/h) on a dry horizontal road, on an icy horizontal road, and on an icy inclined road respectively. So, the stopping distaces on an icy road is 6 times greater and about 47 times greater on a slightly inclined road with respect to the dry horizontal road respectively.