This is too complicated to set up here. draw the figure, Remember, at the wall and floor, you can have a normal and a along surface force.
Now, Sum moments about one end of the ladder and set those equal to zero.
It ought to work out very quickly.
A ladder of unknown length and mass of 29 kg rests on a rough floor, and leans against a rough wall. The coefficient of static friction between the ladder and both the floor and wall are 0.68. Hanging on the ladder are two masses. The first mass is 9.9 kg and is located 5% from the ladder's base. The second mass is 7.1 kg and is located 4% from the top of the ladder. What is the smallest angle (in degrees) between the floor and ladder that the ladder can be set at where it and its masses will not slide?
2 answers
I am a Hong Kong student. But i think i may help you. The only force on the ladder from the wall is the horizontal force.The system here is in static equilibrium. Net torque is zero.Let the total length of the ladder be 1m.
then,
(1)[(29+9.9+7.1)(g)(0.68)]sin⊙
-(0.96)[(7.1)g]cos⊙
-(0.5)[(29)g]cos⊙
-(0.05)[(9.9)g]cos⊙=0
Solving it and you should find the answer(⊙ is the angle)
then,
(1)[(29+9.9+7.1)(g)(0.68)]sin⊙
-(0.96)[(7.1)g]cos⊙
-(0.5)[(29)g]cos⊙
-(0.05)[(9.9)g]cos⊙=0
Solving it and you should find the answer(⊙ is the angle)
1 answer