A uniform ladder AB of wieght W and length 2l rests with the end A against a smooth vertical wall and the end B on a rough horizontal ground. A man of weight equal to that of the ladder stands at the point C on the ladder, where BC = 5l/3 . The coefficient of friction between the ladder and the ground is 1/3. Given that the ladder is in limiting equilibrium when it makes an angle s to the horizontal;
(a) Show that tans = 2
(b) Find the magnitude of the reaction force at the wall and on the ground.
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Expert's answer
2021-02-08T18:42:15-0500
let O middle of ladder
AO=OB=l;AC=3l;BC=35l
s−angle s to the horizontal;
Ffr−friction force at point B
TA−wall reaction force
TB−reaction force at point B
mg−the gravity of the ladder at a point O
mg−the gravity of the man at a point C
forces along the X-axis (horizontal)
TA−Ffr=0
forces along the Y-axis (vertical)
TB−mg−mg=0;TB−2mg=0;
equilibrium condition about point B
TAsin(s)−mg∗cos(s)∗AC−mg∗cos(s)∗AO=0
TAsin(s)−mg∗cos(s)∗31−mg∗cos(s)∗l=0
TA=34cot(s)mg
Ffr=34cot(s)mg
a)friction static of the ladder and the man−31mg+31mg=32mg
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