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A student is standing on a ladder as shown in the figure to the right. Each leg of the ladder is 2.6 m long and is hidged at point C. The tie-rod (BD) attached halfway up and is 0.7 m long. The student is standing at a spot 1.95 m along the leg and her weight is 541 N. (You may ignore the weight of the ladder and any minor friction between the floor and the legs.)

1)

What is the tension in the tie-rod?

N

2)

What is the vertical component of the force of the ground on the ladder leg at point A?

N

3)

What is the horizontal component of the force of the ground on the ladder leg at point A?

N

4)

What is the vertical component of the force of the ground on the ladder leg at point E?

N

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Answer #1

If the tie-rod is "halfway up and is 0.7 m long," then A is 2*0.70m = 1.4 m from E.
The angle each leg makes with horizontal is
Θ = arccos(0.7/2.6) = 74.4º

4) Sum the moments about A:
ΣM = 0 = E*1.4m - 541N * 1.95m * cos74.4º
E = 283.7N·m / 1.4m = 202.6 N

3) If we "may ignore ... any minor friction," then the horizontal component of the force at A (and E) is "minor." If friction is zero, then that component is zero.

2) Sum the vertical forces:
ΣF = 0 = A + E - 541N
A = 541N - E = 338.4 N

1) Cut the ladder in half vertically and sum the moments about C (using the left half):
ΣM = 0 = 541N * (2.6 - 1.95)m * cos74.4º + T * ½ * 2.6m * sin74.4º - 338.4 * 0.7m
T = 113.7 N

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