Question

Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of${\mathbf{34}}^{\mathbf{o}}$. (a) What is the minimum value of the coefficient of static friction that will keep the snow from sliding down? (b) As the snow begins to melt, the coefficient of static friction decreases and the snow finally snips. Assuming that the distance from the chunk to the edge of the roof is 4.0 m and the coefficient of the kinetic friction is 0.10, calculate the speed of the snow chunk when it slides off the roof. (c) If the roof edge is 10.0 m above ground, estimate the speed of the snow when it hits the ground.

Short answer

(a) The minimum value of the coefficient of static friction is 0.67.

(b) The speed of the chunk when it reaches the edge of the roof is $5.76\mathrm{m}/\mathrm{s}$.

(c) The speed of the snow chunk just before hitting the ground is $14.56\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Step 1. Understanding the direction of action of frictional force

Frictional force acts in a direction that enables it to minimize the relative motion between two surfaces.

Step 2. Given data and assumptions

The angle of the slope of the roof,$\theta =34°$

The coefficient of kinetic friction,${\mu }_{\mathrm{k}}=0.10$

The distance between the initial positions of the chunk and the edge of the roof,$l=4.0\mathrm{m}$

The height of the roof edge,$h=10\mathrm{m}$

Let the minimum value of the coefficient of static friction required to stop ice from slipping down be ${\left({\mu }_{\mathrm{s}}\right)}_{\min }$.

Let the acceleration of the snow chunk along the roof, when it starts moving, be a and the velocity of the edge of the roof be v.

Step 3. Calculating the minimum value of the coefficient of static friction

The FBD of the chunk of snow, when it is stationary on the roof, can be drawn as:

In the above figure, the following are provided:

mg – gravitational pull by the earth on snow

N – Reaction force on snow due to roof

f_s – The force of static friction

Applying the condition of equilibrium along the direction perpendicular to the incline, you get:

$N=mg\cos \theta \dots \left(i\right)$

Applying the condition of equilibrium along the incline, you get:

$f=mg\sin 34°$

As static friction is less than or equal to the maximum value of static friction, thus,

${\left(f_{\mathrm{s}}\right)}_{\max }\ge mg\sin 34°$

By the definition of the maximum possible value of static friction,

${\mu }_{\mathrm{s}}N\ge mg\sin 34°$

From (i),

${\mu }_{\mathrm{s}}mg\cos 34°\ge mg\sin 34°$

For finding ${\left({\mu }_{\mathrm{s}}\right)}_{\min }$,

${\left({\mu }_{\mathrm{s}}\right)}_{\min }mg\cos 34°\ge mg\sin 34°$

From the above equation,

$\begin{array}{c}{\left({\mu }_{\mathrm{s}}\right)}_{\min }=\tan 34°\\ =0.67\end{array}$

Thus, the minimum value of the coefficient of friction will be 0.67.

Step 4. Calculating the value of the velocity as it reaches the end of the roof

According to Newton’s second law, acceleration in the direction along the incline can be written as:

$mg\sin \theta -f=ma$

Substituting the values in the above equation, you get:

$mg\sin 34°-{\mu }_{\mathrm{k}}N=ma$

From equation (i),

$mg\sin 34°-0.10\times mg\cos 34°=ma$

Solving the above equation for a, you get:

$\begin{array}{c}a=0.42\times 9.8\\ =4.15\mathrm{m}/{\mathrm{s}}^2\end{array}$

Using the third equation for motion along the roof, you get:

$v^2-0=2al$

Substituting the values in the above equation, you get:

$v^2=2\times 4.15\times 4$

From the above equation,

$v=5.76\mathrm{m}/\mathrm{s}$

Thus, the velocity of the chunk when it is at the edge of the roof is $5.76\mathrm{m}/\mathrm{s}$.

Step 5. Calculating the velocity just before hitting the ground

The velocity in the horizontal direction as the chunk leaves the roof is given as:

$\begin{array}{c}{v}_{\mathrm{x}}=v\cos 34°\\ =5.76\cos 34°\\ =4.96\mathrm{m}/\mathrm{s}\end{array}$

Since there is no acceleration in the horizontal direction, the horizontal component of velocity will not change with time.

The velocity in the vertical direction as the chunk leaves the roof is:

$\begin{array}{c}{v}_{\mathrm{y}}=v\sin 34°\\ =5.76\sin 34°\\ =2.93\mathrm{m}/\mathrm{s}\end{array}$

The acceleration in the vertical direction is equal to the acceleration due to gravity. Thus, by the third equation of motion, the final velocity in the vertical direction can be written as:

${\left(v_{\mathrm{y}}\right)}_{\mathrm{f}}^2-v_{\mathrm{y}}^2=2gh$

Substituting the values in the above equation, you get:

${\left(v_{\mathrm{y}}\right)}_{\mathrm{f}}^2=2\times 9.8\backslash \mathrm{kern}1pt\times 10-2.{93}^2$

Solving the above equation, you get:

${\left(v_{\mathrm{y}}\right)}_{\mathrm{f}}=13.69\mathrm{m}/\mathrm{s}$

The magnitude of the final velocity $v_{\mathrm{f}}$, just before the snow chunk hits the ground, can be written by the formula of the magnitude of a vector as:

$\begin{array}{c}{v}_{\mathrm{f}}=\sqrt{{\left(v_{\mathrm{y}}\right)}_{\mathrm{f}}^2+v_{\mathrm{x}}^2}\\ =\sqrt{{13.69}^2+{4.96}^2}\\ =14.56\mathrm{m}/\mathrm{s}\end{array}$

Thus, the final velocity of the chunk just before hitting the ground is $14.56\mathrm{m}/\mathrm{s}$.