Question

a uniform ladder rests against a rough wall so that it makes an angle of 60.0 degrees with the ground. the ladder is 10.0m long and weighs 150N. How far can a 250N man go before the ladder slips? The coefficient of friction between the ladder and the ground is 0.400; between the ladder and the wall is 0.450.

Answer 1

A uniform ladder rests against a rough wall so that it makes an angle of 60.0 degrees with the ground. the ladder is 10.0m long and weighs 150N. How far can a 250N man go before the ladder slips? The coefficient of friction between the ladder and the ground is 0.400; between the ladder and the wall is 0.450.

Solution:

$\mu_{1} = 0.400 -$ coefficient of friction between the ladder and the ground ;

$\mu_{2} = 0.450 -$ coefficient of friction between the ladder and the wall;

$\alpha = 60{}^{\circ}$ angle which ladder makes with the ground

$N_{1}$ - reaction force from the ground

$N_{2}$ - reaction force from the wall

$P_{1} = 150N - \text{weight of the ladder}$

$P_{2} = 250N - \text{weight of the man}$

$L = 10.0m - \text{length of the ladder}$

We will consider the extreme case when the person is standing at a maximum distance $d$ from the beginning of the ladder.

Newton's second law for the ladder (the first law of equilibrium):

$$\overrightarrow {F _ {f r 1}} + \overrightarrow {F _ {f r 2}} + \overrightarrow {P _ {1}} + \overrightarrow {P _ {2}} + \overrightarrow {N _ {1}} + \overrightarrow {N _ {2}} = \vec {0}$$

Projection of the law on the X-axis:

$$x: N _ {2} - F _ {f r 1} = 0$$

Projection of the law on the X-axis:

$$y: N _ {1} + F _ {f r 2} - P _ {1} - P _ {2} = 0$$

Law of dry friction:

$$F _ {f r 1} = \mu_ {1} N _ {1}$$ $$F _ {f r 2} = \mu_ {2} N _ {2}$$

(3) and (4) in (1) and (2):

$$(3) \rightarrow (1): N _ {2} - \mu_ {1} N _ {1} = 0$$ $$N _ {1} = \frac {N _ {2}}{\mu_ {1}}$$ $$(4) \rightarrow (2): N _ {1} + \mu_ {2} N _ {2} - P _ {1} - P _ {2} = 0$$ $$(5) \rightarrow (6): \frac {N _ {2}}{\mu_ {1}} + \mu_ {2} N _ {2} - P _ {1} - P _ {2} = 0$$ $$N _ {2} = \frac {\mu_ {1} (P _ {1} + P _ {2})}{1 + \mu_ {1} \mu_ {2}} = \frac {0 . 4 0 0 \cdot (1 5 0 N + 2 5 0 N)}{1 + 0 . 4 0 0 \cdot 0 . 4 5 0} = 1 3 5. 6 N$$ $$F _ {f r 2} = \mu_ {2} N _ {2} = 0. 4 5 0 \cdot 1 3 5. 6 N = 6 1. 0 2 N$$

Momentum equation for point A (the second law of equilibrium):

$$A: M _ {P _ {1}} + M _ {P _ {2}} + M _ {f r 2} + M _ {N _ {2}} = 0$$

$(M_{N_1} = M_{fr1} = 0, \text{because moment arm of this forces is zero})$

$$M _ {P _ {1}} = - P _ {1} \cdot \frac {L}{2} \cos \alpha \text{ (minus sign because of the direction of force)}$$ $$M _ {P _ {2}} = - P _ {2} \cdot d \cos \alpha$$ $$M _ {f r 2} = F _ {f r 2} \cdot L \cos \alpha$$ $$M _ {N _ {2}} = N _ {2} \cdot L \sin \alpha$$ $$\rightarrow (6):$$ $$F _ {f r 2} \cdot L \cos \alpha + N _ {2} \cdot L \sin \alpha - P _ {2} \cdot d \cos \alpha - P _ {1} \cdot \frac {L}{2} \cos \alpha = 0$$ $$d = \frac {F _ {f r 2} \cdot L \cos \alpha + N _ {2} \cdot L \sin \alpha - P _ {1} \cdot \frac {L}{2} \cos \alpha}{P _ {2} \cos \alpha} =$$ $$= \frac {6 1 . 0 2 N \cdot 1 0 m \cdot \frac {1}{2} + 1 3 5 . 6 N \cdot 1 0 m \cdot \frac {\sqrt {3}}{2} - 1 5 0 N \cdot \frac {1 0 m}{2} \cdot \frac {1}{2}}{2 5 0 N \cdot 0 . 5} = 8. 8 3 5 m$$

Answer: man can go up the stairs a distance of $d = 8.835m$ before the ladder slips.