The uniform 10-kg ladder in the figure rests against the smooth wall at B, and the end A rests on the rough horizontal plane for which the coefficient of static friction is 0.3. determine the angle of inclination of the ladder and the normal reaction at b if the ladder is on the verge of slipping. The ladder is 4m long.
16,7
To determine the angle of inclination and the normal reaction at point B when the ladder is on the verge of slipping, we need to consider the forces acting on the ladder.
Let's break down the forces involved:
1. Weight (W): It acts vertically downward from the center of mass of the ladder, which we can assume to be at its midpoint. The weight can be calculated as W = m * g, where m is the mass of the ladder and g is the acceleration due to gravity (9.8 m/s^2).
Given that the ladder has a mass of 10 kg, the weight of the ladder is W = 10 kg * 9.8 m/s^2 = 98 N.
2. Normal Reaction (N): It acts perpendicular to the horizontal plane at point A, preventing the ladder from sinking into the surface. The normal reaction also acts perpendicular to the wall at point B. Since the ladder is about to slip, the maximum normal reaction occurs at point B when the ladder is on the verge of slipping.
3. Frictional Force (f): It acts parallel to the horizontal plane in the direction opposite to the impending motion. The frictional force can be calculated as f = μ * N, where μ is the coefficient of static friction.
Given that the coefficient of static friction is 0.3, we can calculate the maximum frictional force as f = 0.3 * N.
To find the angle of inclination, we need to analyze the equilibrium of forces in both the vertical and horizontal directions:
In the vertical direction:
Sum of forces = 0
N - W = 0
N = W
In the horizontal direction:
Sum of forces = 0
f = 0.3 * N
Since N = W = 98 N, the maximum frictional force is f = 0.3 * 98 N = 29.4 N.
To find the angle of inclination (θ), we'll use the formula:
sin(θ) = opposite/hypotenuse
sin(θ) = f/W
θ = arcsin(f/W)
Substituting the values, we get:
θ = arcsin(29.4 N/98 N)
θ ≈ 17.15 degrees
Therefore, the angle of inclination of the ladder is approximately 17.15 degrees.
To determine the angle of inclination of the ladder and the normal reaction at point B, we can analyze the forces acting on the ladder.
1. Draw a diagram of the forces acting on the ladder.
- The ladder exerts a normal force (N) at point B, perpendicular to the wall.
- The weight of the ladder (mg) acts downwards, vertically.
- There is static friction (f) acting at the base of the ladder, resisting its tendency to slip.
2. Resolve the forces vertically and horizontally.
- Vertically: N - mg = 0 (since the ladder is not accelerating vertically)
Therefore, N = mg.
3. Determine the force of static friction.
- The maximum static friction force (f_max) is given by f_max = coefficient of static friction * N
f_max = (0.3) * mg
4. Find the angle of inclination of the ladder.
- Since the ladder is on the verge of slipping, the force of static friction must be at its maximum.
- The force of static friction acts horizontally and can be resolved into horizontal and vertical components.
- The horizontal component should balance the horizontal component of the weight of the ladder.
f_max * cos(theta) = mg * sin(theta)
Dividing by mg and rearranging, we get:
f_max/mg = sin(theta)/cos(theta)
tan(theta) = f_max/mg
5. Calculate the value of tan(theta).
- Substitute the value of f_max from step 3.
tan(theta) = (0.3) * mg / mg
tan(theta) = 0.3
6. Find the angle of inclination, theta.
- Use the inverse tangent function to find theta.
theta = atan(0.3)
theta ≈ 16.69°
7. Calculate the normal reaction at point B.
- Use the equation for the vertical forces, N - mg = 0, which we found in step 2.
N = mg
N = 10kg * 9.8 m/s² (Assuming g = 9.8 m/s²)
N = 98N
Therefore, the angle of inclination of the ladder is approximately 16.69°, and the normal reaction at point B is 98N.