uniform ladder 25 ft long rests against a smooth vertical wall. The ladder weighs 15 Kg. The lower end of the ladder is 15 ft from the wall. A man weighing 80 Kg climbs up the ladder until he is 20 ft from the base of the ladder, (20 ft of the ladder length not above the floor). At this point the ladder starts to slip. What is the coefficient of friction between the ladder and the floor?
see this figure:
http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PS9/9-63.JPG
when the ladder slips, friction Fv*mu is just equal to the force horizontal.
Now sketch the figure on paper,
1) sum the moments about the base and set to zero.
2) sum horizontal forces and set to zero
3) sum vertical forces and set to zero.
You should have three equations, three unknowns, and can solve for mu.
To find the coefficient of friction between the ladder and the floor, we need to analyze the forces acting on the ladder.
First, let's determine the forces acting on the ladder before it starts to slip:
1. Weight of the ladder (15 Kg):
The weight acts vertically downward, and its magnitude is given by W_ladder = m_ladder * g, where m_ladder is the mass of the ladder (15 Kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
W_ladder = 15 Kg * 9.8 m/s^2 = 147 N
2. Normal force on the ladder:
The ladder rests on the floor, causing the floor to exert an upward normal force on the ladder. The magnitude of this normal force is equal to the weight of the ladder.
Normal force = W_ladder = 147 N
3. Friction force:
The ladder is on the verge of slipping, which means the friction force is at its maximum value (maximum static friction). The magnitude of the static friction force is given by F_friction = coefficient of friction * Normal force.
F_friction = coefficient of friction * Normal force
Now, let's analyze the forces acting on the ladder when it starts to slip:
1. Weight of the ladder and the man (95 Kg):
The weight acts vertically downward, and its magnitude is given by W_total = (m_ladder + m_man) * g, where m_man is the mass of the man (80 Kg), m_ladder is the mass of the ladder (15 Kg), and g is the acceleration due to gravity (approximately 9.8 m/s^2).
W_total = (15 Kg + 80 Kg) * 9.8 m/s^2 = 931 N
2. Normal force on the ladder:
The ladder and the man exert a greater force on the floor due to their combined weight, causing an increased normal force. The magnitude of this normal force is equal to the total weight of the ladder and the man.
Normal force = W_total = 931 N
3. Friction force:
When the ladder starts to slip, the friction force decreases to the kinetic friction force. The magnitude of the kinetic friction force is given by F_friction = coefficient of friction * Normal force.
F_friction = coefficient of friction * Normal force
Since the maximum static friction force is equal to the kinetic friction force, we can equate the two friction forces and solve for the coefficient of friction:
coefficient of friction * Normal force = coefficient of friction * Normal force
F_friction = F_friction
So, the coefficient of friction between the ladder and the floor can be found by calculating the ratio of the weight of the ladder to the normal force:
coefficient of friction = F_friction / Normal force = (W_ladder / Normal force) / (W_total / Normal force)
coefficient of friction = (W_ladder / W_total)
Plugging in the given values:
coefficient of friction = (147 N / 931 N)
coefficient of friction ≈ 0.158
Therefore, the coefficient of friction between the ladder and the floor is approximately 0.158.