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 pitch of 21.
What is the minimum value of the coefficient of static friction that will keep the snow from sliding down in units uK?
Me don't know
To solve this problem, we can analyze the forces acting on the chunk of snow at the ridge of the roof.
First, let's consider the gravitational force acting on the chunk of snow. The magnitude of this force can be calculated using the formula:
F_gravity = m * g
Where:
- F_gravity is the gravitational force
- m is the mass of the snow chunk
- g is the acceleration due to gravity (approximately 9.8 m/s²)
Next, let's consider the frictional force that prevents the snow from sliding down the roof. This force depends on the coefficient of static friction (μ) and the inclination of the roof (θ).
The magnitude of the frictional force can be calculated using the formula:
F_friction = μ * N
Where:
- F_friction is the frictional force
- μ is the coefficient of static friction
- N is the normal force, which is equal to the gravitational force component perpendicular to the roof surface. In this case, it is given by:
N = m * g * cos(θ)
Now, to find the minimum value of the coefficient of static friction (μ_min) that will keep the snow from sliding down, we need to set the frictional force equal to the gravitational force. So, we have:
F_friction = F_gravity
μ * N = m * g
Substituting the expression for N:
μ * (m * g * cos(θ)) = m * g
Simplifying:
μ * cos(θ) = 1
μ_min = 1 / cos(θ)
Where:
- θ is the angle of inclination of the roof (given as 21°)
Now, let's calculate the minimum value of the coefficient of static friction (μ_min):
μ_min = 1 / cos(21°)
Calculating this using a scientific calculator:
μ_min ≈ 1.0509
So, the minimum value of the coefficient of static friction that will keep the snow from sliding down is approximately 1.0509 units uK.
To determine the minimum value of the coefficient of static friction that will keep the snow from sliding down the roof, we can use the concept of static equilibrium.
In this case, we consider the chunk of snow as being in equilibrium, meaning that the sum of the forces acting on it in the vertical direction must be zero. The forces involved are the weight of the snow and the frictional force opposing the potential sliding motion.
The weight of the snow can be calculated using the formula W = mg, where m is the mass of the snow and g is the acceleration due to gravity. The mass of the snow can be obtained by multiplying its volume by its density.
The frictional force can be determined using the equation Ff = μsN, where μs is the coefficient of static friction and N is the normal force acting perpendicular to the roof. The normal force is equal to the weight of the snow.
Given that the roof has a pitch of 21, we can use trigonometry to determine the component of the weight of the snow acting parallel to the roof. The normal force is then equal to the weight of the snow times the cosine of the angle of the roof pitch.
To find the minimum value of the coefficient of static friction, we need to set up the equilibrium equation, considering the forces acting vertically. This equation can be written as:
mg = μsN.
Substituting the expressions for the weight of the snow (mg) and the normal force (weight of the snow * cos(21)), we get:
m * g = μs * (weight of the snow * cos(21)).
Simplifying the equation, we can cancel out the mass and divide both sides by the weight of the snow:
g = μs * cos(21).
Finally, we can solve for the minimum value of the coefficient of static friction (μs) in units uK by dividing both sides of the equation by the value of the acceleration due to gravity (g):
μs = (g / cos(21)).
Calculating this expression will give you the minimum value of the coefficient of static friction required to keep the snow from sliding down the roof.