Auniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. the ladder weighs 160 n. the coefficient of static friction between the foot of the ladder and the ground is 0.40. a man weighing 740 n climbs slowly up the ladder. start by drawing a free-body diagram of the ladder.
(a) what is the maximum friction force that the ground can exert on the ladder at its lower end?
(b) what is the actual friction force when the man has climbed 1.0 m along the ladder?
(c) how far along the ladder can the man climb before the ladder starts to slip?

The maximum friction force, actual friction and distance of the ladder are respectively

Formula for finding friction is given as

F = μN, or in this case, F = μ(N1 + N2)

F = 0.40(740 + 160)

F = 0.40(900)

F = 360 N

Look at the attached image, we can agree that from the triangle,

cosθ =

sinθ =

if so, then the moment about the ground point

740 * 1 * cosθ + 2.5 * 160 * cosθ - 5 * n * sinθ

740 * 0.6 + 2.5 * 160 * 0.6 - 5n * 0.8

444 + 240 - 4n

684 - 4n = 0

4n = 684

n = 684/4

n = 171 N

For the last part,

740 * x * cosθ + 2.5 * 160 * cosθ - 5 * n * sinθ

If the maximum friction force, n is 360, then

740x * 0.6 + 2.5 * 160 * 0.6 - 5 * 360 * 0.8

444x + 240 - 1440

444x - 1200

444x = 1200

x = 1200/444

x = 2.7m

to read more about kinematics of motion, see link

(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force =

=360 N

(b)Moment about Ground Point

(c)

Here maximum friction force can be 360 N

Therefore

Where x is the maximum distance moved by man along the ladder

740x=2000

x=2.702m

learn it on the intertant so you can solve

it breaks and dirupture other materials and other sencews from the bottom