Question

A ladder of mass M and length 4 m rests against a frictionless wall at an angle of 50 degrees to the horizontal. The coefficient of static friction between the ladder and the floor is 0.65. What is the maximum distance along the ladder a person of mass 15M can climb before the ladder starts to slip?

Answer 1

Answer on Question #56953, Physics / Mechanics | Relativity

A ladder of mass M and length 4 m rests against a frictionless wall at an angle of 50 degrees to the horizontal. The coefficient of static friction between the ladder and the floor is 0.65. What is the maximum distance along the ladder a person of mass 15M can climb before the ladder starts to slip?

Solution:

We know at the point of slipping that $F_{Gx}$ (See diagram above) is just equal to the friction force, where the normal force is $F_{Gy}$ (The upward force exerted by the ground on the ladder - see diagram above.)

So, applying the formula for friction:

$F_{fr} = m_s F_n$ (at the point of slipping)

where $m_{s}$ is the coefficient of static friction between the ladder and the floor

$$F _ {G x} = m _ {s} F _ {G y}$$

Vertical Force:

Ladder weight: $-Mg$ (down)

Person weight: $-15Mg$ (down)

Ground pushing up: $+F_{Gy}$ (up)

Equilibrium:

$$F _ {G y} - M g - 1 5 M g = 0$$ $$F _ {G y} - 1 6 M g = 0$$

Thus,

$$F _ {G y} = 1 6 M g$$

Horizontal Force:

Force exerted by the wall: $-F_{W}$ (left)

Force exerted by the ground horizontally: $+F_{Gx}$ (right)

Equilibrium:

$$F _ {G x} - F _ {W} = 0$$ $$F _ {G x} = F _ {W} = 0. 6 5 \cdot 1 6 M g = 1 0. 4 M g$$

And finally torque:

We then have the following torques about the bottom of the ladder:

The ground: torque = 0 (r = 0)

The weight of the ladder: $Mg$ N at an angle of $50{}^{\circ}$ at a distance of $4/2 = 2$ m from the bottom of the ladder. $Torque_{1} = (2.0 \, m)(Mg \, N)\cos(50{}^{\circ})$ (CW)

The weight of the person on the ladder: $15\mathrm{Mg}$ N at an angle of $50{}^{\circ}$ at a distance of $x$ m from the bottom of the ladder. $Torque_{2} = x \cdot 15Mg \cdot \cos(50{}^{\circ})$ (CW)

The wall pushing to the left: $F_{W}$ acting at a distance of $4.0 \, \text{m}$ from the bottom, at an angle of $50{}^{\circ}$ with the ladder.

$$Torque_{3} = -(4.0 \, m) F_{W} \sin(50{}^{\circ}) \quad \text{(ACW)}$$

Equilibrium:

$$\begin{aligned} Torque_{1} + Torque_{2} + Torque_{3} &= 0 \\ 2Mg\cos(50{}^{\circ}) + 15xMg\cos(50{}^{\circ}) - 4 \cdot 10.4Mg\sin(50{}^{\circ}) &= 0 \\ 2\cos(50{}^{\circ}) + 15x\cos(50{}^{\circ}) - 41.6\sin(50{}^{\circ}) &= 0 \\ x &= \frac{41.6 \sin(50{}^{\circ}) - 2\cos(50{}^{\circ})}{15\cos(50{}^{\circ})} = 1.311 \, \text{m} \end{aligned}$$

Answer: 1.311 m

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