Question

A 1,071.628 N painter needs to climb d=1.926 m up a ladder (measured along its length from the point where the ladder contacting the ground), without the ladder slipping. The uniform ladder is 12.014 m long and weighs 250 N. It rests with one end on the ground and the other end against a perfectly smooth vertical wall. The ladder rises at an angle of theta=51.96 degrees above the horizontal floor. What is friction force in unit of N that the floor must exert on the ladder? Use g = 10 m/s2 if you need to .

Answer 1

The frictional force in unit of N that the floor must exert on the ladder is approximately 232.216 N

The known values are;

The weight of the painter = 1,071.628 N

The height to which the painter needs to climb along the ladder = 1.926 m

The length of the ladder = 12.014 m

The weight of the ladder = 250 N

The points where one of the ladder's ends is resting = On the ground

The points where the other end of the ladder is resting = A perfectly smooth wall

The angle with which the ladder rises above the horizontal floor = 51.96°

The acceleration due to gravity, g ≈ 10 m/s²

The unknown values include;

The friction force that the floor must exert on the ladder

The strategy to be used;

At equilibrium, the sum of moments about a point is zero

Finding the moments about the point of contact where the ladder rests on the wall, P, is given as follows;

At equilibrium, the sum of clockwise, $M_{CW}$, moment about P = The sum of the counterclockwise, $M_{CCW}$moment about P

$\mathbf{M_{CCW}}$ = (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

$\mathbf{M_{CW}}$ = 12.014 × cos(51.96°) × $\mathbf{F_N}$

Where;

$\mathbf{F_N}$ = The normal reaction of the of the ground on the end of the ladder that rests on the floor

$\mathbf{M_{CCW}}$ = $\mathbf{M_{CW}}$

∴ (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250 = 12.014 × cos(51.96°) × $F_N$

We get;

6,665.3068846 N·m =  7.40316448688 m × $F_N$

$\mathbf{F_N}$ = 6,665.3068846 N·m/(7.40316448688 m) = 900.332135 N

The normal reaction of the floor on the ladder, $\mathbf{F_N}$ = 900.332135 N

Taking moment about the point the ladder rests on the floor, R, gives;

$M_{CCW}$ = 12.014 × sin(51.96°) × $F_W$

Where;

$\mathbf{F_W}$ = The normal reaction at the wall

$M_{CW}$ = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

At equilibrium, we have, $M_{CCW}$ = $M_{CW}$

Therefore;

12.014 × sin(51.96°) × $F_W$ = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

9.46199511627 m × $F_W$ = 2,197.22861125 N·m

$F_W$ = 2,197.22861125 N·m/(9.46199511627 m)

The reaction of the wall, $\mathbf{F_W}$ = 232.216206 N

We note that also at equilibrium, the sum horizontal forces = 0

The horizontal forces acting  on the ladder = The normal reaction on the, $F_W$ wall and the friction force on the ground, $\mathbf{F_f}$

∴ At equilibrium; $\mathbf{F_W}$ + $\mathbf{F_f}$ = 0

$\mathbf{F_f}$ = -$\mathbf{F_W}$

$\mathbf{F_W}$  = 232.216206 N

Therefore;

The frictional force in unit of N that the floor must exert on the ladder, $\mathbf{F_f}$ = 232.216206 N ≈ 232.216 N.

(The coefficient of friction, μ = $\mathbf{F_N}$/$\mathbf{F_W}$ = 900.332135/232.216206 ≈ 3.877).

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