A uniform ladder of length 20.0m and weight 750 N is propped up against a smooth vertical wall with its lower end on a rough horizontal surface. The coefficient of friction between the ladder and this horizontal surface is 0.40.
(b) Work out and add the numerical values of each force clearly showing your justification in each case.
(c) Hence, calculate a value for the angle between the ladder and the wall if the ladder just remains in stable equilibrium.
So now sum vertical forces: You only have two (weightladder and floor upward).
Sum horizontal foces: wall force and friction at base.
Sum moments about any point: I recommend the floor contact point, you have ladder weight, and wall force then.
The angle should come put pretty quickly on the moment equation.
I will be happy to check your calcuations.
weight ladder plus floor upward: 750 + 20.0= 770N
wall force + friction at base: 20.0 + 0.40= 20.40N
I have no idea what you mean about sum moments about any point.
Wow, you have missed some class time.
Try this: http://info.ee.surrey.ac.uk/Teaching/Courses/ee1.el3/s07el3.pdf
That doesn't help me because I still don't understand it. Please can you at least direct me on how to go about the question as my work is due very soon.
To solve this problem, we need to analyze the forces acting on the ladder in order to determine the angle between the ladder and the wall when it is in stable equilibrium.
First, let's consider the forces acting on the ladder:
1. Weight (W): This force is acting vertically downward and has a magnitude of 750 N.
2. Normal force from the ground (N1): This force is perpendicular to the ground and acts upward to support the ladder. Its magnitude will be the same as the weight of the ladder, which is 750 N.
3. Normal force from the wall (N2): This force is perpendicular to the wall and acts in a direction opposite to the weight of the ladder. Its magnitude will also be 750 N.
4. Frictional force (F_friction): This force acts parallel to the ground and opposes the tendency of the ladder to slide. Its magnitude can be determined using the coefficient of friction and the normal force from the ground. We can calculate it as F_friction = coefficient of friction * N1 = 0.40 * 750 N = 300 N.
5. Tension force (T): This force acts along the ladder and provides stability. Its direction is opposite to the weight of the ladder. Its magnitude can be found by considering the vertical and horizontal components of the force.
Now, let's resolve the forces acting along the ladder into their vertical and horizontal components.
1. Vertical Forces: The vertical components of the weight of the ladder, normal force from the ground, and tension force should balance each other. We can write the equation as follows:
N1 - W * sin θ - T * cos θ = 0 (Equation 1)
Here, θ represents the angle between the ladder and the wall.
2. Horizontal Forces: The horizontal components of the weight of the ladder and the frictional force should balance each other. We can write the equation as follows:
W * cos θ - F_friction = 0 (Equation 2)
From Equation 2, we can solve for the value of θ:
W * cos θ - F_friction = 0
750 N * cos θ - 300 N = 0
Simplifying the equation:
cos θ = 300 N / 750 N
θ = cos^(-1) (0.4)
Using a calculator, we find:
θ ≈ 66.42 degrees
Therefore, the angle between the ladder and the wall when the ladder is in stable equilibrium is approximately 66.42 degrees.