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3 years ago

8

An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground

is 0.55, and the ladder makes a 50.0° angle with the ground. How far up the ladder can an 710 N person climb before the ladder begins to slip?

1 answer:

sweet [91]3 years ago

7 0

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m

AC = distance of person on the ladder from the bottom end = x

W = weight of the ladder = 240 N

$F_{g}$ = weight of the person = 710 N

F = force by the wall on the ladder

N = normal force by ground on the ladder = ?

Using equilibrium of force along the vertical direction

N = $F_{g}$ + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f =static frictional force on the ladder

Static frictional force is given as

f = μ N

f = (0.55) (950)

f = 522.5 N

Force equation along the horizontal direction is given as

F = f

F = 522.5 N

using equilibrium of torque about point A

F Sin50 (AD) = W Cos50 (AB) + ( $F_{g}$ Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

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$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

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We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

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$| \vec{L}_{rod_f} | = r_f * m * v_f$

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$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

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$v_f = 15 \frac{m}{s}$

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