Consider the stability of a ladder of mass m and length L leaning against a wall. A painter wants to be sure the ladder will not slip across the floor. He knows that the frictional force exerted by the floor on the bottom end of the ladder is needed to keep the ladder in equilibrium, and he also knows that the coefficient of static friction between the ladder and the floor is μSf = 0.80. Assume there is friction between the vertical wall and the ladder, with μSw = 0.28.
Find the angle at which the ladder just begins to slip.
To find the angle at which the ladder just begins to slip, we need to consider the forces acting on the ladder.
Let's break down the forces acting on the ladder:
1. The weight of the ladder, acting at the center of mass and directed downward. This force can be represented as m * g, where m is the mass of the ladder and g is the acceleration due to gravity.
2. The normal force exerted by the floor on the ladder, acting perpendicular to the floor and directed upward.
3. The frictional force between the ladder and the floor, acting parallel to the floor and opposing the motion of the ladder.
4. The frictional force between the ladder and the wall, acting parallel to the wall and opposing the motion of the ladder.
At the point of impending slip, the frictional force exerted by the floor on the bottom end of the ladder is at its maximum value, which is equal to the product of the normal force and the coefficient of static friction between the ladder and the floor, i.e., μSf * (m * g).
Taking the sum of the forces in the horizontal direction, we have:
Frictional force between the ladder and the floor = Frictional force between the ladder and the wall.
μSf * (m * g) = μSw * (m * g)
Canceling out the mass and acceleration due to gravity, we get:
μSf = μSw
Substituting the given values, we have:
0.80 = 0.28
Since the equation is not satisfied, it means that the ladder will not start to slip. Therefore, there is no specific angle at which the ladder just begins to slip in this scenario.
To find the angle at which the ladder just begins to slip, we can start by analyzing the forces acting on the ladder.
First, let's consider the forces acting on the ladder in the horizontal direction. The vertical wall exerts a normal force on the ladder, which is perpendicular to the wall and cancels out the gravitational force pulling the ladder downwards. The frictional force between the ladder and the wall opposes the tendency of the ladder to slide down. Since the ladder is not slipping horizontally, the frictional force between the wall and the ladder must be equal to the horizontal component of the force due to gravity.
Next, let's consider the forces acting on the ladder in the vertical direction. The normal force of the floor on the bottom end of the ladder equalizes the gravitational force acting on it in the vertical direction.
Now, let's calculate the forces involved. The force due to gravity acting on the ladder can be calculated as Fg = mg, where m is the mass of the ladder and g is the acceleration due to gravity. The normal force exerted by the floor on the bottom end of the ladder is equal to the gravitational force, so N = mg.
To find the maximum frictional force between the ladder and the floor, we multiply the coefficient of static friction (μSf) by the normal force exerted by the floor:
Ff = μSf * N
Similarly, the maximum frictional force between the wall and the ladder can be calculated as:
Fw = μSw * N
Since we know that the frictional force between the wall and the ladder must be equal to the horizontal component of the force due to gravity, we have:
Fw = Fg * sinθ
where θ is the angle between the ladder and the horizontal.
Now, let's set up equations using the calculated quantities:
Ff = μSf * N
Fw = Fg * sinθ
Substituting in the values we know:
μSf * N = μSw * N
mg * sinθ = μSw * mg
Canceling out the mass and rearranging the equation:
sinθ = μSw
Finally, to find the angle at which the ladder just begins to slip, we can take the inverse sine of both sides of the equation:
θ = arcsin(μSw)
Plugging in the given value for μSw:
θ = arcsin(0.28)
Using a calculator, we can find:
θ ≈ 16.92 degrees.
Therefore, the ladder just begins to slip at an angle of approximately 16.92 degrees.