In the figure, a uniform ladder of weight 200 N and length 10 m leans against a perfectly smooth wall. A firefighter of weight 600 N climbs a distance x up the ladder. The coefficient of static friction between the ladder and the floor is 0.50. What is the maximum value of x for which the ladder will not slip?
X being the hypotenuse from the ground to the fireman
To solve this problem, we need to consider the forces acting on the ladder and the equilibrium condition that must be satisfied for it to not slip.
Let's analyze the forces involved:
1. Weight of the ladder (200 N): This force acts downward and can be considered to act at the center of mass of the ladder, which is at its midpoint.
2. Weight of the firefighter (600 N): This force acts downward and can be considered to act at the center of mass of the firefighter, which is at his midpoint.
3. Normal force (N): This force acts perpendicular to the ground and is equal to the sum of the ladder's weight and the firefighter's weight. It can be calculated as: N = 200 N + 600 N = 800 N.
4. Frictional force (F_friction): This force acts parallel to the ground and opposes the motion or tendency of motion between the ladder and the floor. The maximum value of frictional force is given by the product of the coefficient of static friction (µ) and the normal force as: F_friction = µN.
Now, let's consider the equilibrium condition:
For the ladder to be in equilibrium and not slip, the sum of the forces acting along the ladder (vertical direction) and along the ground (horizontal direction) must be zero.
Along the vertical direction (y-axis):
Sum of forces = 0
N - Weight of the ladder - Weight of the firefighter = 0
N - 200 N - 600 N = 0
N = 800 N.
Along the horizontal direction (x-axis):
Sum of forces = 0
F_friction - Tension in the ladder = 0
µN - Tension = 0
Since the ladder does not slip, the tension in the ladder is also equal to the weight of the firefighter.
Tension in the ladder = Weight of the firefighter = 600 N.
Substituting the values of N and the tension in the above equation, we get:
µ*(800 N) - 600 N = 0
µ = 600 N / (800 N)
µ = 0.75
We have the maximum value of the coefficient of static friction as 0.75.
Now, to find the maximum value of x, we can use the tangent of the angle between the ladder and the ground.
Tangent of the angle = Opposite / Adjacent
x / 10 m = 600 N / 200 N
x = (600 N / 200 N) * 10 m
x = 30 m
Therefore, the maximum value of x for which the ladder will not slip is 30 meters.
To find the maximum value of x for which the ladder will not slip, we need to consider the different forces acting on the ladder.
Let's break down the forces acting on the ladder:
1. Weight of the ladder (represented by W_ladder) = 200 N, acting vertically downward from the center of mass of the ladder.
2. Weight of the firefighter (represented by W_firefighter) = 600 N, acting vertically downward from the center of mass of the firefighter.
3. Normal force exerted by the floor (represented by N) = equal and opposite to the sum of W_ladder and W_firefighter, acting vertically upward.
4. Force of static friction between the ladder and the floor (represented by f_friction) = depends on the coefficient of static friction (μ) and is equal in magnitude to μ * N.
In this case, the ladder will not slip as long as the force of static friction is greater than or equal to the horizontal component of the weight of the ladder. This is because the static friction provides the necessary opposing force to prevent the ladder from sliding down the wall.
Now, let's find the value of the horizontal component of the weight of the ladder:
The ladder forms a right triangle with the wall and the floor, where x represents the hypotenuse from the ground to the firefighter. The angle between the ladder and the floor can be determined by using trigonometry.
In a right triangle, the cosine of an angle (θ) can be defined as the adjacent side divided by the hypotenuse. In this case, the adjacent side is the horizontal component of the weight of the ladder, and the hypotenuse is x.
cos(θ) = adjacent side / hypotenuse
cos(θ) = W_ladder_horizontal / x
W_ladder_horizontal = x * cos(θ)
Since we know that cos(θ) = adjacent side / hypotenuse, we can rearrange the equation to solve for cos(θ):
cos(θ) = (W_ladder_horizontal) / (W_ladder)
W_ladder_horizontal = W_ladder * cos(θ)
In this case, the weight of the ladder (W_ladder) is given as 200 N, and we can calculate the cosine of θ based on the value of x.
Next, we need to determine the force of static friction (f_friction) acting on the ladder:
f_friction = μ * N
Since N is equal to the sum of W_ladder and W_firefighter:
N = W_ladder + W_firefighter
And since W_ladder_horizontal = W_ladder * cos(θ), we can substitute the value of N and W_ladder_horizontal into the equation:
f_friction = μ * (W_ladder + W_firefighter)
Now, we set up the inequality condition:
f_friction ≥ W_ladder_horizontal
μ * (W_ladder + W_firefighter) ≥ W_ladder * cos(θ)
Substituting the given values, we have:
0.50 * (200 N + 600 N) ≥ 200 N * cos(θ)
Simplifying the equation:
400 N ≥ 200 N * cos(θ)
Now, we need to solve for θ. We can do this by taking the inverse cosine (also known as the arccosine) of both sides of the equation:
cos^(-1)(400 N / (200 N)) ≥ θ
cos^(-1)(2) ≥ θ
Using a calculator, we find that θ is approximately 60 degrees.
Finally, we can find the maximum value of x by considering the right triangle formed by the ladder, the wall, and the floor, and using the tangent function:
tan(θ) = opposite side / adjacent side
tan(60 degrees) = x / W_ladder_horizontal
Substituting the values:
tan(60 degrees) = x / (200 N * cos(60 degrees))
Using a calculator, we find that the maximum value of x is approximately 3.08 meters.
Therefore, the maximum value of x for which the ladder will not slip is approximately 3.08 meters.