Question

A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800-N firefighter has climbed 4.00 m along the ladder from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 9.00 m from the bottom, what is the coefficient of static friction between ladder and ground

Answer 1

a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36

Step-by-step explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

-w x - W x₂ + R y = 0 (1)

usemso trigonometry to find distances

cos 60.08 = x / 7.5

x = 7.5 cos 60.08

x = 3.74 m

fireman

cos 60.08 = x₂ / 4

x2 = 4 cos 60

x2 = 2 m

wall support

sin 60.08 = y / 15

y = 15 are 60.08

y = 13 m

we substitute in equation 1

R y = w x + W x2

R = (w x + W x2) / y

R = (500 3.74 +800 2) / 13

R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

R -fr = 0

R = fr

fr = 266.92 N

Y Axis

Fy - w-W = 0

fy = 500 + 800

fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

cos 60.08 = x₃ / 9.00

x₃ = 9 cos 60

x₃ = 5.28 m

from equation 1

R = (w x + W x₃) / y

R = 500 3.74 + 800 5.28) / 13

R = 468.769 N

we saw that

fr = R = 468.769

The expression for the friction force is

fr = μ N

in this case the normal is the ratio to pesos

N = Fy

N = 1300 N

μ N = fr

μ = fr / N

μ = 468,769 / 1300

μ = 0.36