In Figure 12-6, a ladder of length L = 12 m and mass m = 46 kg leans against a slick (frictionless) wall. The ladder's upper end is at height h = 9.2 m above the pavement on which the lower end rests (the pavement is not frictionless). The ladder's center of mass is L/3 from the lower end. A firefighter of mass M = 65 kg climbs the ladder. Let the coefficient of static friction μs between the ladder and the pavement be 0.45. How far (in percent) up the ladder muct the firefighter go to put the ladder on the verge of sliding?
Don't you have any questions without missing figures?
To determine how far up the ladder the firefighter must go to put the ladder on the verge of sliding, we need to analyze the forces acting on the ladder.
1. Start by drawing a free-body diagram of the ladder, considering the forces acting on it.
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F_f Fr
F_f = force of friction between the ladder and the pavement
Fr = normal force exerted by the wall on the ladder
mg = weight of the ladder
2. Consider the torque equation for equilibrium about the bottom end of the ladder (point O).
τ = (L/3) * Fr - (2L/3) * mg = 0
Since the ladder is in equilibrium, the net torque about point O is zero.
3. The friction force is given by:
F_f = μs * N
N = mg - Fr
Fr can be determined using trigonometry:
sin(θ) = h / L
Fr = mg * sin(θ)
4. Substitute the expressions for Fr and N into the torque equation:
(L/3) * mg * sin(θ) - (2L/3) * mg = 0
5. Solve for sin(θ):
sin(θ) = (2/3)
6. Calculate the angle θ:
θ = sin^(-1)(2/3)
7. To find the vertical displacement (d) when the firefighter reaches the critical point, we can use the following equation:
d = L - h
8. Calculate the vertical displacement:
d = L - h = 12 - 9.2 = 2.8 m
9. Finally, calculate the percentage of the ladder's length the firefighter must climb to reach the critical point:
percentage = (d / L) * 100 = (2.8 / 12) * 100 = 23.3%
Therefore, the firefighter must climb approximately 23.3% up the ladder to put it on the verge of sliding.
To find out how far up the ladder the firefighter must go to put the ladder on the verge of sliding, we need to analyze the forces acting on the ladder.
First, let's calculate the gravitational force acting on the ladder:
F_gravity = mass x acceleration due to gravity
F_gravity = m x g
F_gravity = 46 kg x 9.8 m/s^2
F_gravity = 450.8 N
Next, let's determine the normal force exerted by the pavement on the ladder. Since the ladder is on the verge of sliding, we know that the static friction is at its maximum value. The normal force is equal to the gravitational force because the ladder is not accelerating vertically:
F_normal = F_gravity
F_normal = 450.8 N
The maximum static friction force is given by the equation:
F_friction = μs x F_normal
Plugging in the values:
F_friction = 0.45 x 450.8 N
F_friction = 202.86 N
Now, let's consider the forces acting horizontally. When the ladder is on the verge of sliding, the force of friction is equal to the force exerted by the firefighter. So:
F_friction = force exerted by the firefighter
Let's assume the firefighter exerts a force F_firefighter while climbing up the ladder. The force exerted by the firefighter can be expressed as:
F_firefighter = M x acceleration due to gravity
F_firefighter = 65 kg x 9.8 m/s^2
F_firefighter = 637 N
Now, we can set up an equation to find the distance the firefighter must go up the ladder to put it on the verge of sliding:
F_friction = F_firefighter
μs x F_normal = F_firefighter
Solving for the distance the firefighter must climb up the ladder:
μs x F_normal = M x g x distance climbed
0.45 x 450.8 N = 65 kg x 9.8 m/s^2 x distance climbed
0.45 x 450.8 N / (65 kg x 9.8 m/s^2) = distance climbed
0.45 x 450.8 N / 637 N = distance climbed
0.31936 ≈ distance climbed
Finally, to express the distance climbed as a percentage of the ladder's length L:
distance climbed / L x 100% = percentage distance climbed
0.31936 / 12 m x 100% = 2.6613 ≈ 2.7%
Therefore, the firefighter must go approximately 2.7% up the ladder to put it on the verge of sliding.