Question

In the figure, a ladder of length L = 10 m and mass m = 44 kg leans against a slick (frictionless) wall. The ladder's upper end is at height h = 7.9 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 = 68 kg climbs the ladder. Let the coefficient of static friction μs between the ladder and the pavement be 0.51. How far (in percent) up the ladder must the firefighter go to put the ladder on the verge of sliding?

Answer 1

Ans:-

Given data: L= 10m, m= 44kg, h= 7.9m, M= 68kg,μs= 0.51

Since there is no friction at the wall, the normal force on the ground is the sum of the weights of the ladder and the firefighter:
Fn = (44 + 68)kg * 9.8m/s² = 1097.6 N
so the maximum friction force is
Ff = µ*Fn = 0.51 * 1097.6N = 559.78N

The angle between the ground and the ladder is
Θ = arcsin(7.9/10) = 52.19º

Now sum the moments about the base:
ΣM = 0 = firefighter's torque + ladder's torque - wall force's torque
0 = 68kg*9.8m/s²*x*10m*cos52.19 + 44kg*9.8m/s²*(10m/3)*cos52.19 - Fw*7.9m
where Fw is the horizontal force at the wall which, at the threshold, is equal to the friction force at the floor (the only two horizontal forces on the ladder).
0 =4085.33Nm*x +881.15Nm -4422.26Nm

0= 4085.33*x Nm -3541.11Nm
x = 3541.11N·m / 4085.33N·m = 0.8667 = 0.87=87%