Question

An 8.0 m, 280 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is 0.65, and the ladder makes a 50.0° angle with the ground. How far up the ladder can an 870 N person climb before the ladder begins to slip?

Answer 1

Let:
S be the horizontal reaction of the wall on the top of the ladder,
W1 be the weight of the ladder,
W2 be the weight of the person,
R be the vertical reaction of the ground on the foot of the ladder,
F be the friction force towards the wall at the foot of the ladder,
a be the inclination of the ladder to the ground,
x be the distance the person can climb up the ladder,
2L be the length of the ladder,
u be the coefficient of static friction between the ladder and the ground.

Resolving vertically and horizontally:
W1 + W2 = R ...(1)
F = S ...(2)

F / R <= u ...(3)

Moments about the foot of the ladder:
(W1 L + W2 x)cos(a) = 2SL sin(a)

W2 x cos(a) = 2SL sin(a) - W1 L cos(a)
2SL tan(a) = W1 L + W2 x

Using (2) to eliminate S:
2FL tan(a) = W1 L + W2 x
F = (W1 L + W2 x) / [ 2L tan(a) ] ...(4)

Eliminating R from (1) and (3):
F <= u(W1 + W2) ...(5)

Using (5) to eliminate F from (4):
(W1 L + W2 x) / [ 2L tan(a) ] <= u(W1 + W2)

W1 L + W2 x <= 2uL(W1 + W2) tan(a)
W2 x <= 2uL(W1 + W2) tan(a) - W1 L
x <= (L / W2)[ 2u(W1 + W2)tan(a) - W1 ]

x <= (4.0 / 790)[ 2 * 0.7(280 + 790)tan(50) - 280 ]
x <= 7.62 m to 3 sig. fig.