Question

If the coefficient of static friction is 0.753, the length of the ladder is 9.9 m, and its mass is 39 kg, find the minimum height LaTeX: h below which the ladder will slip.

Answer 1

The ladder will slip at the point where the reaction at the wall is just over

the force due to friction.

Response:

• The minimum height below which the ladder will slip, is approximately 5.48 meters above the ground.

Which method is used to calculate the minimum height before slipping?

The given parameter are;

The coefficient of friction, μ = 0.753

Length of the ladder = 9.9 m

Mass of the ladder = 39 kg

Required:

The minimum height below which the ladder will slip.

Solution:

Assumption: The friction of the wall on the ladder is 0.

The weight of the ladder, W = The normal reaction = N

The friction force, $F_f$ = The reaction force of the wall, $F_w$

$F_f$ = W × μ

Which gives;

$F_f$ = 39 × 9.81  × 0.753 ≈ 288.09

The friction force, $F_f$ ≈ 288.09 N = $F_w$

Taking moments about the contact between the ladder and the ground, we have;

$F_w$ × h = W × x

Where;

$h = \mathbf{L \times sin(\theta)}$

$x = \mathbf{\dfrac{L}{2} \times cos(\theta)}$

Which gives;

$x = \dfrac{9.9}{2} \times cos(\theta) = \mathbf{4.95 \cdot cos(\theta)}$

$h = 9.9 \cdot sin(\theta)$

θ = The angle made by the ladder and the ground

Therefore;

288.09 × 9.9·sin(θ) =  39 × 9.81 × x  = 382.59 × 4.95·cos(θ)

$\dfrac{sin(\theta)}{cos(\theta)} = tan(\theta) = \dfrac{382.59 \times 4.95}{288.09 \times 9.9}$

$\theta = arctan \left(\dfrac{382.59 \times 4.95}{288.09 \times 9.9} \right) \app$

Which gives;

$h = \mathbf{9.9 * sin\left(arctan \left(\dfrac{382.59 \times 4.95}{288.09 \times 9.9} \right) \right)} \approx 5.48$

The minimum height below which the ladder will slip, h ≈ 5.48 m

Learn more about friction here:

brainly.com/question/13879636