Question

A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 65.0° angle with the horizontal.

(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.20 m along the ladder from the bottom. Horizontal Force Magnitude = Direction = Vertical Force Magnitude = Direction =
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?

Answer 1

we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.

Coefficient of static friction = 0.26

Step-by-step explanation:

Given that:

length of the ladder = 16.0 m

weight of the ladder = 520 N

angle θ = 65.0°

(a) We are to find the horizontal and vertical forces the ground exerts on the base of the ladder when an :

Force = 850 N

distance of the climber from the base of the ladder = 4.20 m

The diagrammatic illustration representing what the given information entails can be seen from the attached file below.

Let consider the Ladder being at point A with the horizontal layer of the ground.

From the whole system; the condition for the equilibrium at the point A can be computed as :

`N_2 (16 \ Sin\ 65) = 850(4.2 \ * Cos \ 65 )+ 520 ((16)/(2)) Cos \ 65`

`N_2 (14.50) = 850(1.7749 )+ 520 (8) * 0.4226`

`N_2 (14.50) = 1508.665+1758.016`

`N_2 (14.50) = 3266.681`

`N_2 =( 3266.681)/(14.50)`

`N_2 =225.28 \ N`

`N_1 = mg+F\\`

where ;

w =mg

`N_1 = 520+850`

`N_1 = 1370 \ N`

Therefore; we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.

(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?

the coefficient of static friction between ladder and ground when the firefighter is 9.40 m from the bottom can be calculated as:

`N_2 (16 \ Sin\ 65) = 850(9.4 \ * Cos \ 65 )+ 520 ((16)/(2)) Cos \ 65`

`N_2 (14.50) = 850(3.9726 )+ 520 (8) * 0.4226`

`N_2 (14.50) =3376.71+1758.016`

`N_2 (14.50) =5134.726`

`N_2 =(5134.726)/(14.50)`

`N_2 =354.12 \ N`

Therefore; the coefficient of the static friction is;

`\mu = (f_s)/(N_1)`

`\mu = (354.12)/(1370)`

`\mu` = 0.26

Coefficient of static friction = 0.26