Asked by Anonymous

An 8.5 m extension ladder of mass 24 kg is propped up against a wall, touching at a point 8.0 m above the level ground. A man of mass 75 kg climbs 2.0 m up the ladder to repair a window. The ladder rests against a frictionless wall, but the ground has friction. Determine the magnitude and direction of all forces on the ladder.

Normal force exerted by the ground.
(Answer: 970N)

Frictional force exerted by ground.
(Answer: 104N)

Force exerted by wall.
(Answer: 104N)

Force exerted by man.
(Answer: 735N)

Force exerted by weight of ladder.
(Answer: 235N)

Answers are provided above. Please show all work on how to get the answers.
Answered by Anonymous
a) 24kg * 9.8m/s^2 + 75kg * 9.8m/s^2 = 970N

b) A = sqrt[(8.5m)^2 - (8m)^2] = 2.87m
CC = 2m
CC / AC = 2m/B = 8m / 2.87m
AC= 2m * 2.87 / 8m
AC = 0.7175m

BA = 8.5m / 2m = 4.25m

AC / BA = AO / AD
AC / 4.25m = 2.87m / 8.5m
AC2 = 2.87m * 4.25m / 8.5m
AC2 = 1.435m

F2AC1 + F1AC2 - Ff * 8m = 0
Ff = 75kg * 9.8m/s^2 * 2.5m + 24kg * 9.8m/s^2 * 0.41m / 8m
Ff = (735N * 0.7175m + 235.2N * 1.435m) / 8m
Ff = 104N

c) Force exerted by wall = Frictional force exerted by ground.
Force by wall = 104NN

d) Force of man = 75kg * 9.8m/s^2 = 735N

e) Force of ladder = 24kg * 9.8m/s^2 = 235N
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