- #1
mistermill
- 19
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A weightless ladder 7.0 m long rests against a frictionless wall at an angle of 65° ablove the horizontal. A 72 kg person is 1.2m from the top of the lass. What horizontal force at the bottom of the ladder is required to keep it from slipping?
Στ=0
τ=FsinθdSolutions for this question always have the pivot point at the bottom of the ladder, solving for the horizontal frictional force at the bottom of the ladder as being equal and opposite to the force at the top.Στ=0
τ person = mgsin25°(5.8)
τ friction at the top= Fsin65°(7.0)
F = 272N
My question is WHY doesn't the answer work if we place the pivot point at the top?
Στ=0
τ person = mgsin25°(1.2)
τ friction at the bottom = Fsin65°(7.0)
F = 56N
What is missing in this solution that would make it correct?
Στ=0
τ=FsinθdSolutions for this question always have the pivot point at the bottom of the ladder, solving for the horizontal frictional force at the bottom of the ladder as being equal and opposite to the force at the top.Στ=0
τ person = mgsin25°(5.8)
τ friction at the top= Fsin65°(7.0)
F = 272N
My question is WHY doesn't the answer work if we place the pivot point at the top?
Στ=0
τ person = mgsin25°(1.2)
τ friction at the bottom = Fsin65°(7.0)
F = 56N
What is missing in this solution that would make it correct?