A uniform ladder of mass, m, and length, l, leans at an angle, theta, against a rough wall and is on a rough floor. Determine a formula for the minimum angle at which the ladder will not slip if a man reach the top of the ladder.
θ=42°
To determine the minimum angle at which the ladder will not slip, let's analyze the forces acting on the ladder. We need to consider the normal force (N) exerted by the floor on the ladder, the normal force (Nw) exerted by the wall on the ladder, the weight force (mg) acting on the ladder's center of mass, and the friction forces (Ff) between the ladder and the floor and wall.
For the ladder to not slip, the friction forces must be sufficient to prevent the ladder from sliding either down the wall or across the floor. The friction force is given by the equation Ff = μN, where μ is the coefficient of friction.
Now let's analyze the forces in the vertical direction. The normal force exerted by the floor, N, can be decomposed into a component perpendicular to the floor (Ncosθ) and a component parallel to the floor (Nsinθ). The normal force exerted by the wall, Nw, acts perpendicular to the wall.
The weight force, mg, acts vertically downward through the center of mass of the ladder. The vertical forces are balanced, so we have the equation Ncosθ + Nw = mg.
Next, let's consider the forces in the horizontal direction. The friction forces, Ff, act in the opposite direction to the motion that could cause the ladder to slip. For the ladder not to slip down, the friction force at the floor must be greater than or equal to 0. Similarly, for the ladder not to slip sideways, the friction force at the wall must be greater than or equal to 0.
The magnitude of the friction force between the ladder and the floor is given by Ff(floor) = μ(Ncosθ). The magnitude of the friction force between the ladder and the wall is given by Ff(wall) = μ(Nsinθ).
Now, we can set up the inequalities for the friction forces: Ff(floor) ≥ 0 and Ff(wall) ≥ 0.
μ(Ncosθ) ≥ 0 (1)
μ(Nsinθ) ≥ 0 (2)
Since μ > 0 and Ncosθ ≥ 0 and Nsinθ ≥ 0, we can conclude that equation (1) is always satisfied.
Solving equation (2) for θ, we get:
Nsinθ ≥ 0
sinθ ≥ 0
The sine function is greater than or equal to 0 in the range -π/2 ≤ θ ≤ π/2. Therefore, for the ladder not to slip, the minimum angle at which the ladder will not slip is θ = -π/2 or θ = π/2.
Since the ladder cannot lean downward, the minimum angle at which the ladder will not slip is θ = π/2.
Therefore, the formula for the minimum angle is θ = π/2.