Question

A ladder of length L = 2.3 m and mass m = 14 kg rests on a floor with coefficient of static friction μ_s = 0.54. Assume the wall is frictionless.

A person with mass M = 69 kg now stands at the very top of the ladder.

What is the normal force the floor exerts on the ladder?

What is the minimum angle to keep the ladder from sliding?

Answer 1

Starting off with the Free Body Diagram depicting all forces acting on the ladder

Now Friction always opposes the motion of an object or if the object is at rest it opposes the tendency of the object to move. In this case the ladder has a tendency to move downward due to gravity therefore the lower half of the ladder will tend to move towards the right as shown in second figure

This implies that the tendency of motion is towards the left, hence friction will act on that point in the opposite direction towards the right (refer Force Diagram).

Now we are required to find Normal reaction exerted by the floor and the minimum angle for the rod to not slide.

First we assume θ to be the minimum angle at which the rod is at rest (not slipping).

To Find Normal Reaction :-

Since bosy is at rest, the algebraic sum of all forces acting on the body is zero

Sum of forces in horizontal direction is zero

Sum of forces acting towards left = sum of forces acting towards right

N₂ = Fr.

Also, sum of forces in vertical direction is zero

Sum of upward forces = sum of downward forces

N₁ = mg + Mg......... eqn. 1

Given m=14 & M=69 and assuming g=0.98

N₁ = (14 x 9.8) + (69 x 9.8) = 813.4 N (newtons).......(required Normal reaction of floor)

To Find Minimum Angle :-

Firstly we know that     

in this case .....eqn. 2 ..................( Given mu = 0.54 and N = N₁= 813.4 newtons)

Now to calculate Fr -

We know since body is at rest the algebraic sum of rotational moments about any point is zero.

Sum of clockwise moments = Sum of anticlockwise moments

Considering moments about point of ladder in contact with the wall. (refer diagram)

Since N₂ & Mg act about the point of rotation their moment is zero.

sum of clockwise moments = N₁ x (L x cosθ)

sum of anticlockwise moments = mg x (L/2 x cosθ) + Fr x (L x sinθ)

N₁ x (L x cosθ) = mg x (L/2 x cosθ) + Fr x (L x sinθ)

cancelling L throughout and solvinf for Fr -

Fr x sinθ = ( N₁ + mg/2) cosθ

Fr = (813.4 + 68.6) cotθ

equating Fr with eqn. 2 :-

882 cotθ ≤ 439.2

tanθ ≥ 882/439.2 = 2.01

θ ≥ tan^-1(2.01) = 63.55^°

minimum θ = 63.55^°