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=
unanswered
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=
unanswered
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)
I don't know the rest.
a) fs = gcotanè(mp/3+ml/2)
To solve this problem, we need to analyze the forces acting on the person and the ladder. Let's start by drawing a free-body force diagram for each.
1. Free-Body Diagram for the Person:
- The person is acted upon by their weight (mg) downward and the contact force exerted by the ladder (Cperson,ladder) upward.
- Since the person is standing on a rung, there is no horizontal force acting on them.
2. Free-Body Diagram for the Ladder:
- The ladder is acted upon by its weight (m1g) downward, the normal force exerted by the wall (Nwall) perpendicular to the wall, the normal force exerted by the floor (Nfloor) perpendicular to the floor, and the contact force exerted by the person (Cperson,ladder) parallel to the floor.
- The ladder experiences torque around its bottom end due to the normal force exerted by the floor.
Now, let's proceed with solving the questions:
(a) Finding the magnitude of the frictional force (fs) between the base of the ladder and the floor below:
- Since the ladder is in static equilibrium, the net force and net torque acting on it must be zero.
- The horizontal force equation: Cperson,ladder - fs = 0 (Since there is no horizontal acceleration)
- The torque equation: (d/3) * (m1g) - (2d/3) * fs - [(2d/3) - d] * Cperson,ladder = 0 (Taking the bottom end of the ladder as the pivot point)
Solving these two equations simultaneously will give us the value of fs.
(b) Finding the magnitude of the minimum coefficient of friction (μs):
- To prevent the ladder from slipping, the frictional force must be greater than or equal to the minimum value required to overcome the tendency to slip.
- μs = fs / Nfloor
(c) Finding the magnitude of Cperson,ladder,ground:
- The contact force exerted by the floor on the ladder is the vector sum of the normal force (Nfloor) and the frictional force (fs).
- Cperson,ladder,ground = sqrt(Nfloor^2 + fs^2)
To find the direction of the contact force, we need to determine the angle α that the contact force makes with the horizontal.
- α = atan(fs / Nfloor)
By solving the equations and substituting the given values, you can find the answers to the questions (a), (b), (c), and the direction α.