A uniform ladder of length 2a and mass m lies on a vertical plane with one end against a smooth vertical wall,the other end being supported on a horizontal floor. The ladder is released from rest when inclined at an angle Q to the horizontal. Find the inclination of the ladder to the horizontal when it ceases to touch the wall.
To find the inclination of the ladder to the horizontal when it ceases to touch the wall, we need to analyze the forces acting on the ladder.
Let's consider the ladder in equilibrium just before it loses contact with the wall. At that point, the normal force at the point of contact with the floor is zero.
Now, let's break down the forces acting on the ladder:
1. Weight: The weight of the ladder acts vertically downward. Since the ladder is uniform, its weight can be considered to act at its center of mass, which is at the midpoint. The magnitude of the weight is mg, where m is the mass of the ladder and g is the acceleration due to gravity.
2. Normal force at the point of contact with the floor (Nf): Since the ladder is in equilibrium, the normal force must balance the weight. Therefore, Nf = mg.
3. Normal force at the point of contact with the wall (Nw): The normal force at the point of contact with the wall is perpendicular to the wall and acts horizontally. This force prevents the ladder from sliding down. Let's call the angle between the ladder and the horizontal θ. The vertical component of Nw is Nwsin(θ), and it balances the weight of the ladder. Therefore, Nwsin(θ) = mg.
Now, we can determine the relationship between θ and Q. Let's consider the forces acting on the ladder in the horizontal direction:
1. Friction force (f): The friction force acts on the ladder at the point of contact with the floor. Since the ladder is in equilibrium, the friction force must balance the horizontal component of Nw. Therefore, f = Nwcos(θ).
2. Tension force (T): The tension force acts on the ladder at the point of contact with the wall. It provides the necessary centripetal force to keep the ladder in equilibrium. The tension force can be expressed as T = 2a × f, where 2a is the length of the ladder.
Now, we can relate Q, θ, and the forces acting on the ladder:
In the vertical direction:
Nwsin(θ) = mg
In the horizontal direction:
f = Nwcos(θ)
Substituting Nw = mg/sin(θ) from the vertical direction equation into the horizontal direction equation, we get:
f = mg/tan(θ)cos(θ) = mg/sin(2θ)
Substituting f = T / (2a), we can solve for θ:
mg/sin(2θ) = T / (2a)
Rearranging the equation:
T = 2amg/sin(2θ)
When the ladder ceases to touch the wall, T = 0. Therefore, we have:
0 = 2amg/sin(2θ)
Since sin(2θ) cannot be zero, the only solution is:
2θ = 0
Dividing both sides by 2, we find:
θ = 0
Therefore, the inclination of the ladder to the horizontal when it ceases to touch the wall is 0 degrees. In other words, the ladder becomes completely horizontal.