A uniform ladder 5.10 m long rests against a frictionless, vertical wall with it

Question

A uniform ladder 5.10  m long rests against a frictionless, vertical wall with its lower end 3.00  m from the wall. The ladder weighs 155  N . The coefficient of static friction between the foot of the ladder and the ground is 0.430. A painter weighing 745  N climbs slowly up the ladder

Part A

What is the maximum frictional force that the ground can exert on the ladder at its lower end

Part B

What is the actual size of the frictional force when the painter has climbed 1.1202  m along the ladder

Part C

How far along the ladder can the painter climb before the ladder starts to slip?

Explanation / Answer

Here ,

a)

length of ladder , L = 5.1 m

weight of ladder , Wl = 155 N

distance from wall , d = 3 m

height of ladder , h = sqrt(5.1^2 - 3^2)

h = 4.12 m

let f is the frictional force on the ground

N is the normal force on ground

N1 is the normal force on wall

part A)

normal force , N = weight of ladder + weight of paimter

N = 745 + 155

N = 900 N

maximum frictional force = u * N

maximum frictional force = 0.43 * 900

maximum frictional force = 387 N

the maximum frictional force is 387 N

part B)

for the climber ,

balancing the moment of force about the support on floor

N1 * 4.12 - 155 * (3/2) - 745 * (1.1202/5.1) * 3 = 0

solving for N1

N1 = 176 N

for the horizontal forces

N1 - f = 0

f = N1

f = 176 N

the frictional forces is 176 N

part C)

let the maximum distance is d

normal force , N1 = maximum friction

N1 = 387 N

387 * 4.12 - 155 * (3/2) - 745 * (x/5.1) * 3 = 0

x = 3.11 m

the position where maximum painter can go is 3.11 m

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