A person of mass m2= 85.0 kg is standing on a rung, one third of the way up a ladder of length d= 3.0 m. The mass of the ladder is m1= 15.0 kg, uniformly distributed. The ladder is initially inclined at an angle θ= 35.0∘ with respect to the horizontal. Assume that there is no friction between the ladder and the wall but that there is friction between the base of the ladder and the floor with a coefficient of static friction μs .
Start this problem by drawing a free-body force diagrams showing all the forces acting on the person and the ladder. Indicating a choice of unit vectors on your free-body diagrams may be helpful.
(a) Using the equations of static equilibrium for both forces and torque, find expressions for the normal and horizontal components of the contact force between the ladder and the floor, and the normal force between the ladder and the wall. Consider carefully which point to use for computing the torques. Determine the magnitude of the frictional force (in Newton) between the base of the ladder and the floor below.
fs=
(b) Find the magnitude for the minimum coefficient of friction between the ladder and the floor so that the person and ladder does not slip.
μs=
(c) Find the magnitude Cladder,ground (in Newton) of the contact force that the floor exerts on the ladder. Remember, the contact force is the vector sum of the normal force and friction.
Cladder,ground=
Find the direction of the contact force that the floor exerts on the ladder. i.e. determine the angle α (in radians) that the contact force makes with the horizontal to indicate the direction.
α=
b) Find Mu_s first
Mu_s= (m_p/3 + m_l/2) *cotan (theta)/(m_p+m_l)
a) Force = Mu_s*g(m_p+m_l)
Thanx.. Can you also give the rest?
I already got a and b but for c I thought that by equilibrium C will be the same as a, I'm confused could you give me a hint? Thanks.
To solve this problem, we need to analyze the forces and torques acting on the ladder and the person. By setting up equations of static equilibrium, we can determine the magnitudes and directions of the different forces involved.
(a) Free-body diagram:
Let's draw a free-body diagram for the ladder and the person:
1. Ladder:
- The weight of the ladder acts vertically downward at its center of mass.
- The normal force from the floor acts upward at the contact point between the ladder and the floor.
- The frictional force between the base of the ladder and the floor acts horizontally.
2. Person:
- The weight of the person acts vertically downward at their center of mass.
- The normal force from the ladder acts vertically upward at the contact point between the person and the ladder.
- The force exerted by the wall acts horizontally.
Now, let's set up equations of static equilibrium:
Forces in the vertical direction:
N_ladder - 2*N_person - m1*g - m2*g = 0 ----(1)
Forces in the horizontal direction:
F_wall - f_fric = 0 ----(2)
Torques about the point where the ladder is in contact with the floor:
N_person * (d/3) * cos(θ) - f_fric * (2/3)d * sin(θ) - m1 * g * (d/2) * cos(θ) = 0 ----(3)
From equation (2), we can solve for the frictional force (fs):
f_fric = F_wall
Therefore, fs = F_wall
(b) To find the minimum coefficient of friction (us), we need to consider the tipping point where the ladder just starts to slip. This occurs when the frictional force (fs) reaches its maximum value, which is equal to the product of the normal force (N_ladder) and the coefficient of static friction (us):
fs = us * N_ladder
From equation (1):
N_ladder = 2*N_person + m1*g + m2*g
So, the minimum coefficient of friction (us) is given by:
us = fs / N_ladder = fs / (2*N_person + m1*g + m2*g)
(c) To find the magnitude of the contact force between the ladder and the ground (Cladder,ground), we need to consider the vector sum of the normal force (N_ladder) and the frictional force (fs):
Cladder,ground = sqrt(N_ladder^2 + fs^2)
(d) The direction of the contact force that the floor exerts on the ladder is given by the angle α that the contact force makes with the horizontal. To find α, we can use trigonometry:
tan(α) = fs / N_ladder
α = atan(fs / N_ladder)
Now, we have all the information needed to solve the problem. Plug in the given values for m1, m2, d, θ, and any other known quantities to find the final answers for fs, us, Cladder,ground, and α.