A uniform 7.1 m tall aluminum ladder is leaning against a frictionless vertical wall. The ladder has a weight of 272 N. The ladder slips when it makes a 57.0degrees angle with the horizontal floor.
Determine the coefficient of static friction between the ladder and the floor.
To determine the coefficient of static friction between the ladder and the floor, we need to use the conditions for equilibrium.
First, let's calculate the normal force acting on the ladder:
N = mg + Fsinθ
where:
N = normal force
m = mass of the ladder
g = acceleration due to gravity (9.8 m/s^2)
F = static frictional force
θ = angle made by the ladder with the horizontal floor
We know that the ladder slips when it makes a 57.0° angle with the horizontal floor. So, θ = 57.0°.
Next, we know that the ladder is in equilibrium, so the sum of the forces in the horizontal and vertical directions must equal zero.
In the horizontal direction, the only force acting is the static frictional force, F.
In the vertical direction, we have the weight of the ladder (mg) acting downward and the normal force (N) acting upward.
Since the ladder is in equilibrium, we can write:
ΣF_horizontal = 0
F - Fcosθ = 0
F = Fcosθ
ΣF_vertical = 0
N - mg - Fsinθ = 0
Now, let's substitute the given values:
m = mass of the ladder = weight/g = 272 N / 9.8 m/s^2 = 27.76 kg
θ = 57.0°
g = 9.8 m/s^2
F = Fcosθ
N - mg - Fsinθ = 0
Substituting the values:
N - (27.76 kg)(9.8 m/s^2) - (F)(sin 57.0°) = 0
Solving for N:
N = (27.76 kg)(9.8 m/s^2) + (F)(sin 57.0°)
N = 271.5488 N + (F)(0.8387)
Next, we can substitute this value of N in the equation:
F - Fcosθ = 0
F = Fcosθ
F = (F)(cos 57.0°)
Dividing both sides by F:
1 = cos 57.0°
cos 57.0° = 1
Therefore, we have:
1 = 1
Hence, the equation is true.
So, the coefficient of static friction between the ladder and the floor is 1.
To determine the coefficient of static friction between the ladder and the floor, we need to use the condition for equilibrium in the vertical direction.
In this scenario, the ladder is in equilibrium, meaning the net forces and torques acting on it are balanced. We can consider the forces and torques acting on the ladder to find the coefficient of static friction.
Let's break down the forces acting on the ladder:
1. Weight of the ladder (mg): The weight of the ladder acts downward and can be calculated by multiplying the mass (m) of the ladder by the acceleration due to gravity (g). In this case, the weight is given as 272 N.
2. Normal force (N): The reaction force from the floor acting on the ladder is equal in magnitude and opposite in direction to the weight of the ladder. So, the normal force N is also 272 N.
3. Frictional force (F): The frictional force acts horizontally along the floor and opposes the motion of the ladder. It is equal to the coefficient of static friction (μ) multiplied by the normal force (N).
Now, we can determine the static frictional force:
For the ladder to be in equilibrium, the net vertical force must be zero. This implies that the vertical component of the force acting along the ladder (mg sinθ) must be balanced by the normal force (N):
N = mg sinθ
Substituting the given values:
272 N = m × 9.8 m/s^2 × sin 57°
From here, we can solve for the mass (m) of the ladder. Once we have the mass, we can find the coefficient of static friction (μ) by using the horizontal equilibrium condition.
For the ladder to be in equilibrium horizontally, the net horizontal force must be zero. This implies that the horizontal component of the force acting along the ladder (mg cosθ) must be balanced by the static frictional force (F):
F = μN
We substitute the known values:
F = μ × 272 N
Since the ladder slips right at the point of equilibrium, the static friction force (F) will be at its maximum, given by the maximum static frictional force (Fmax). Therefore:
Fmax = 272 N
Setting Fmax equal to μN:
272 N = μ × 272 N
Now we can solve for the coefficient of static friction (μ):
μ = 272 N / 272 N = 1
So, the coefficient of static friction between the ladder and the floor is equal to 1.
The frictional torque is barely able to keep it from slipping at 57 degrees from horizontal. Set the net torque about the point of contact with the vertical wall equal to zero.
Wladder*(L/2)*cos57 = Wladder*mus*Lsin57
0.5 cot57 = mus = 0.325
"mus" is the static friction coefficient