A woman is standing on the middle of a ladder that leans up against a building, with an angle of 56 degrees from the ground. The wall is very slippery and the ladder is very light. What is the minimum coefficient of friction needed between the ground and the ladder so that the ladder doesn't fall down?
To solve this problem, we can use the concept of equilibrium.
Step 1: Draw a free body diagram of the forces acting on the ladder.
- There are two forces acting on the ladder: the weight of the ladder (W) and the normal force (N) exerted by the ground.
- The weight of the ladder can be split into two components: the vertical component (Wv) and the horizontal component (Wh), where Wh = W * cos(56°) and Wv = W * sin(56°).
- The frictional force (f) also acts horizontally and opposes the horizontal component of the weight.
Step 2: Write the equations for the sum of forces in the vertical and horizontal directions.
- In the vertical direction, the sum of forces is equal to zero (since there is no vertical acceleration): N - Wv = 0.
- In the horizontal direction, the sum of forces is also equal to zero (since there is no horizontal acceleration): f - Wh = 0.
Step 3: Substitute the equations to solve for the minimum coefficient of friction (μ).
- From the equation N - Wv = 0, we can solve for N: N = Wv = W * sin(56°).
- Substituting N into the equation f - Wh = 0, we get: f - Wh = 0.
- Solving for f gives us: f = Wh = W * cos(56°).
Step 4: Substitute the angle of inclination and solve for the minimum coefficient of friction.
- We are given that the angle of inclination is 56 degrees.
- Let's assume the weight of the ladder is W = 1 (to remove the dependence on the actual weight).
- Substitute the values into f = W * cos(56°): f = 1 * cos(56°).
- Calculate the value of f to find the minimum coefficient of friction.
Step 5: Calculate the value of f and the minimum coefficient of friction.
- Using a calculator, find the value of f = cos(56°) = 0.559.
- The minimum coefficient of friction (μ) needed to prevent the ladder from falling down is equal to the calculated value of f.
- Therefore, the minimum coefficient of friction needed is μ = 0.559.
To solve this problem, we can consider the forces acting on the ladder. There are three forces involved: the weight of the ladder pointing downwards, the normal force exerted by the ground on the ladder (perpendicular to the ground), and the frictional force opposing the tendency of the ladder to slip down the wall.
The weight of the ladder can be broken down into two components: one parallel to the wall and one perpendicular to the wall. Given that the ladder is very light, we can ignore the parallel component.
Let's call the normal force N and the coefficient of friction between the ladder and the ground μ. The frictional force is given by F_friction = μN.
Now, considering the forces acting vertically on the ladder, we have:
N - W_perpendicular = 0
Solving for N, we find:
N = W_perpendicular
where W_perpendicular is the perpendicular component of the weight of the ladder.
Now, considering the forces acting horizontally on the ladder, we have:
F_friction = W_parallel
The component of the weight parallel to the wall can be found by multiplying the weight of the ladder by the sine of the angle:
W_parallel = W * sin(θ)
where θ is the angle of the ladder from the ground (56 degrees in this case).
Substituting N = W_perpendicular and F_friction = μN, we can write:
μN = W_parallel
Multiplying both sides by sin(θ) and substituting W_parallel, we get:
μ * W * sin(θ) = W * sin(θ)
Now, we can cancel out the sine of the angle on both sides:
μ = 1
Therefore, the minimum coefficient of friction required between the ground and the ladder to prevent it from slipping is 1.
This means that the ladder does not require any friction at all to remain in place, even on a slippery surface.