A ladder of length L = 2.1 m and mass m = 18 kg rests on a floor with coefficient of static friction μs = 0.57. Assume the wall is frictionless.
a. What is the normal force the floor exerts on the ladder?
b. What is the minimum angle the ladder must make with the floor to not slip?
c. A person with mass M = 66 kg now stands at the very top of the ladder.
What is the normal force the floor exerts on the ladder?
d. What is the minimum angle to keep the ladder from sliding?
a. Well, if you want to know the normal force the floor exerts on the ladder, I suppose you could say it's like a friend who always has your back. In this case, the force is perpendicular to the surface and opposes the weight of the ladder. So, the normal force is equal to the weight of the ladder, which is m * g, where m is the mass of the ladder and g is the acceleration due to gravity. Now you just have to plug in the numbers and solve. Good luck!
b. The minimum angle the ladder must make with the floor to not slip? Hmmm...sounds like we're talking about a ladder that wants to stay put and not go sliding around like a figure skater on ice. So, we need to find the angle at which the force of friction between the ladder and the floor equals the force due to gravity acting parallel to the floor. This force can be found using m * g * sin(θ), where m is the mass of the ladder, g is the acceleration due to gravity, and θ is the angle. Combine that with the force of friction, which is equal to μs * (m * g * cos(θ)), where μs is the coefficient of static friction. Now, set those two forces equal to each other and solve for θ. Trust me, it's a fun mathematical dance.
c. Oh, so now the ladder has a passenger on top? Well, the normal force the floor exerts on the ladder will be the weight of the ladder plus the weight of the person. So, it would be (m + M) * g, where m is the mass of the ladder and M is the mass of the person. Add a little gravity into the mix, and you're good to go!
d. Ah, the minimum angle to keep the ladder from sliding. So, we need to find the angle at which the force of friction between the ladder and the floor reaches its maximum value. This is when the ladder is just about to slide. The force of friction can be calculated using μs * (m * g * cos(θ)), where μs is the coefficient of static friction. Set that equal to the maximum force of friction, which is μs * (m * g) and solve for θ. You got this!
a. To find the normal force the floor exerts on the ladder, we need to consider the forces acting on the ladder in the vertical direction.
The ladder experiences two vertical forces - its weight (mg) acting downward and the normal force (N) exerted by the floor upward. These two forces must balance each other for the ladder to remain in equilibrium.
Since the ladder is at rest, the vertical forces are balanced:
N - mg = 0
Therefore, the normal force exerted by the floor is equal to the weight of the ladder:
N = mg
Given that the mass of the ladder is m = 18 kg, and the acceleration due to gravity is g = 9.8 m/s^2, we can now calculate the normal force:
N = 18 kg * 9.8 m/s^2 = 176.4 N
So, the normal force the floor exerts on the ladder is 176.4 N.
b. To find the minimum angle the ladder must make with the floor to not slip, we need to consider the maximum static friction force between the ladder and the floor.
The maximum static friction force can be calculated using the equation:
F_s = μ_s * N
where μ_s is the coefficient of static friction and N is the normal force.
In this case, the maximum static friction force must counteract the downward force of the ladder's weight (mg), which can be decomposed into its vertical component (mg) and its horizontal component (mg * sinθ).
Thus, to prevent slipping, the maximum static friction force should be equal to the horizontal component of the ladder's weight:
F_s = mg * sinθ
Substituting μ_s * N for F_s, we get:
μ_s * N = mg * sinθ
Rearranging the equation, we can solve for the angle (θ):
sinθ = (μ_s * N) / (mg)
θ = arcsin((μ_s * N) / (mg))
Substituting the given values, where μ_s = 0.57, N = 176.4 N, m = 18 kg, and g = 9.8 m/s^2, we can calculate the minimum angle:
θ = arcsin((0.57 * 176.4 N) / (18 kg * 9.8 m/s^2))
θ ≈ 58.8°
So, the minimum angle the ladder must make with the floor to not slip is approximately 58.8°.
c. When a person stands on the top of the ladder, there are additional forces acting on the ladder. The normal force (N) now needs to support the combined weight of the ladder and the person.
The total weight force (W) acting on the ladder is given by:
W = (m + M) * g
where M is the mass of the person.
To find the normal force, we can set it equal to the total weight force:
N = (m + M) * g
Substituting the given values, where m = 18 kg, M = 66 kg, and g = 9.8 m/s^2, we can calculate the normal force:
N = (18 kg + 66 kg) * 9.8 m/s^2
N ≈ 823.2 N
Therefore, the normal force the floor exerts on the ladder when the person stands at the top is approximately 823.2 N.
d. To find the minimum angle to keep the ladder from sliding, we need to consider both the maximum static friction force and the additional force applied by the person.
The net force acting on the ladder in the horizontal direction must be zero for it not to slide.
The horizontal forces acting on the ladder are the horizontal component of the ladder's weight (mg * sinθ) and the horizontal component of the force applied by the person (Mg).
The maximum static friction force is given by μ_s * N.
Setting the sum of the horizontal components of these forces equal to zero, we have:
mg * sinθ + Mg = μ_s * N
Substituting the values, where m = 18 kg, M = 66 kg, g = 9.8 m/s^2, μ_s = 0.57, and N = 823.2 N, we can solve for the angle (θ):
18 kg * 9.8 m/s^2 * sinθ + 66 kg * 9.8 m/s^2 = 0.57 * 823.2 N
161.4 N * sinθ + 646.8 N = 470.424 N
161.4 N * sinθ = 470.424 N - 646.8 N
161.4 N * sinθ = -176.376 N
sinθ = -176.376 N / 161.4 N ≈ -1.091
However, sine θ cannot be greater than 1, so there is no solution for this equation. This means that it is not possible to prevent the ladder from sliding with the given values.
Therefore, there is no minimum angle to keep the ladder from sliding in this case.
To solve these problems, we need to apply the principles of statics and equilibrium. Let's go through each question step by step:
a. To find the normal force the floor exerts on the ladder when it is just resting, we need to consider the forces acting on the ladder in the vertical direction. The only force present in the vertical direction is the weight of the ladder (mg). The normal force (N) exerted by the floor balances the weight of the ladder, so N = mg. Therefore, the normal force is equal to the weight of the ladder, which is N = (18 kg)(9.8 m/s²) = 176.4 N.
b. To find the minimum angle the ladder must make with the floor to prevent slipping, we need to consider the forces acting on the ladder in both the vertical and horizontal directions. In the vertical direction, we have the weight of the ladder (mg), and in the horizontal direction, we have the frictional force (μsN) opposing the tendency of the ladder to slide. The ladder will not slip when the frictional force just balances the component of weight pulling the ladder downward. So, we have μsN = mg(sinθ), where θ is the angle the ladder makes with the floor. Since we already know N = mg from part a, we can substitute it into the equation above and solve for the angle θ: μs(mg) = mg(sinθ). Simplifying, we get μs = sinθ, so the minimum angle is θ = sin⁻¹(μs) = sin⁻¹(0.57) ≈ 35.6°.
c. When a person stands at the top of the ladder, additional forces are acting on the system. We still have the weight of the ladder (mg) and the normal force (N) exerted by the floor, but now we also have the weight of the person (Mg) acting vertically downward. The normal force has to support both the weight of the ladder and the person. So, the total normal force is N = mg + Mg = (18 kg)(9.8 m/s²) + (66 kg)(9.8 m/s²) = 676.8 N.
d. To find the minimum angle to keep the ladder from sliding when a person is standing at the top, we need to consider the forces in both the vertical and horizontal directions, similar to part b. However, now we also need to take into account the additional downward force caused by the weight of the person (Mg). The ladder will not slide if the frictional force just balances the total component of weight pulling the ladder downward. So, μsN = mg(sinθ) + Mg. Again, substitute N = mg + Mg and solve for the angle θ: μs(mg + Mg) = mg(sinθ) + Mg. Simplifying, we get μs(mg + Mg) = (mg + Mg)(sinθ), and after canceling out the common terms, we have μs = sinθ. Solving for the angle θ, we get θ = sin⁻¹(μs) = sin⁻¹(0.57) ≈ 35.6°.
Therefore, the minimum angle to keep the ladder from sliding remains the same as in part b, regardless of the person standing on top.
a. Fn = M*g = 18 * 9.8 = 176.4 N. = Normal force.
b. Fs = u * Fn = 0.57 * 176.4 = 100.5 N. = force of static friction.
Fp - Fs = M*a.
Fp - 100.5 = 18 * 0,
Fp = 100.5 N. = force parallel with plane to overcome Fs.
Fp = Fn*sinA = 100.5.
176.4*sinA = 100.5,
A = 34.75o = max. angle.