An 84 k g person stands on a uniform ladder 4.0 m long, that weighs 70 N , as sh
ID: 2257079 • Letter: A
Question
An 84kg person stands on a uniform ladder 4.0mlong, that weighs 70N , as shown in the figure(Figure 1) . The floor is rough; hence, it exerts both a normal force, f1, and a frictional force, f2, on the ladder. The wall, on the other hand, is frictionless; it exerts only a normal force, f3.
Explanation / Answer
The forces and torques must be in balance, since nothing is moving. There is an illustration with some limited animation at reference 1.
At the base of the ladder, there are two forces operating:
An upward force normal to the floor F1 supporting the ladder; and
a horizontal frictional force F2 that keeps the ladder from slipping.
At the top of the ladder, there is a force F3 normal to the wall to keep the ladder from moving.
Let:
L = the length of the ladder = 4m
W1 = the weight of the ladder = 70 N
W2 = the weight of the person = 84*g N = 823.2N
? = the angle the ladder makes with the floor
a = the height of the top of the ladder from the floor = 3.8m
b = the distance of the base of the ladder from the wall
L^2 = a^2 + b^2 ==>> b = 1.25
Both horizontal and vertical forces must balance; therefore
(1) F2 = F3 (horizontal)
(2) F1 = W1 + W2 = 70 + 823.2 = 893.2N
The torques about the axis where the ladder meets the floor must also balance:
(3) F3 * L * sin(?) = W1 * 0.5L * cos(?) + W2 * fup * L * cos(?),
where fup = fraction of the way up the ladder where the person is standing
Solving (3) for F3:
(5) F3 = (cos(?) / sin(?)) * (W1 * 0.5 + W2 * fup)
= (1.25 / 3.8) * (70 * 0.5 + 823.2 * fup)
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