Physics College

A 67 kg person climbs up a uniform 12 kg ladder. The ladder is 5 m long; its lower end rests on a rough horizontal floor (static
friction coefficient 0.39) while its upper end rests against a frictionless vertical wall. The angle between the ladder and the horizontal is 43◦.
Let d denote the climbing person’s distance from the bottom of the ladder (see the above diagram). When the person climbs too far (d > dmax), the ladder slips and falls down (kaboom!). Calculate the maximal distance dmax the person will reach before the ladder slips. The
acceleration of gravity is 9.8 m/s*s. Answer in two decimal places max.

A 67 Kg Person Climbs Up A Uniform 12 Kg Ladder. The Ladder Is 5 M Long; Its Lower End Rests On A Rough

Answers

Answer 1

Answer:

1.7 m

Explanation:

Draw a free body diagram of the ladder.  There are 5 forces:

Normal force N pushing up at the base of the ladder.

Friction force Nμ pushing right at the base of the ladder.

Weight force mg pushing down a distance d up the ladder.

Weight force Mg pushing down a distance L/2 up the ladder.

Reaction force R pushing left at the top of the ladder.

Sum of forces in the x direction:

∑F = ma

Nμ − R = 0

Sum of forces in the y direction:

∑F = ma

N − mg − Mg = 0

Sum of moments about the base of the ladder:

∑τ = Iα

mg (d cos θ) + Mg (L/2 cos θ) − R (L sin θ) = 0

Use the first equation to substitute for R:

mg (d cos θ) + Mg (L/2 cos θ) − Nμ (L sin θ) = 0

Use the second equation to substitute for N:

mg (d cos θ) + Mg (L/2 cos θ) − (mg + Mg) μ (L sin θ) = 0

Simplify and solve for d:

m (d cos θ) + M (L/2 cos θ) − (m + M) μ (L sin θ) = 0

m (d cos θ) = (m + M) μ (L sin θ) − M (L/2 cos θ)

d = [ (m + M) μ (L sin θ) − M (L/2 cos θ) ] / (m cos θ)

Plug in values and solve:

d = [ (67 kg + 12 kg) (0.39) (5 m sin 43°) − (12 kg) (2.5 m cos 43°) ] / (67 kg cos 43°)

d = 1.70 m

Rounded to two significant figures, the maximum distance is 1.7 m.

Answer 2

The maximal distance,  [tex]d_{max[/tex], the person will reach before the ladder slips will be 3.41 m.

Here, we need to consider the forces acting and torque equilibrium. Let's start by noting down the given data:

Mass of the person, [tex]m_p[/tex] = 67 kgMass of the ladder, [tex]m_l[/tex] = 12 kgLength of the ladder, L = 5 mCoefficient of static friction, μ = 0.39Angle with the horizontal, θ = 43°

The forces acting on the ladder are:

The weight of the ladder ([tex]W_l[/tex]) acting at its center of mass, which is at L/2.The weight of the person ([tex]W_p[/tex]) acting at distance d from the bottom.The normal force from the ground (N) and the frictional force from the ground (f).The normal force from the wall ([tex]F_w[/tex]), which is horizontal since the wall is frictionless.

Using Newton’s second law for horizontal and vertical equilibrium and setting torques about the base of the ladder:

1. Horizontal forces:

[tex]\[F_w = f\][/tex]

2. Vertical forces:

[tex]N=[/tex] [tex]W_l + W_p[/tex]

For torque equilibrium about the base:

[tex]\[N \times L \sin(\theta) = W_l \left(\frac{L}{2}\right) \cos(\theta) + W_p d \cos(\theta)\][/tex]

Substitute: [tex]\[N = (m_l + m_p)g\][/tex]

And frictional force: f = μN

We solve for d:

[tex]\[0.39(m_l + m_p)g \cdot L \sin(\theta) = m_l g \left(\frac{L}{2}\right) \cos(\theta) + m_p g \cdot d \cos(\theta)\][/tex]

Simplify:

[tex]\[d = \frac{0.39(m_l + m_p)L \sin(\theta) - \left(m_l \frac{L}{2}\right) \cos(\theta)}{m_p \cos(\theta)}\][/tex]

Substituting values:

d = [tex]\[\frac{0.39(12 + 67) \times 5 \sin(43^\circ) - \left(12 \times \frac{5}{2}\right) \cos(43^\circ)}{67 \cos(43^\circ)}\][/tex]d ≈ 3.41 m

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