A 15 M ladder weighing 50 kg rests against a smooth wall at a point 12 m above the ground the centre of gravity of the ladder is one third the above 80 kg man climbs Halfway up the ladder find the forces exerted by the system on the ground and the wall
Answers
Answer :
Hey Mate,
Suppose,
The angle made by the ladder with ground be angle θ.
cos θ = (12/15)
cos θ = (3/5 )
So,
sin θ = 45
#Figure is attach with this answer below. The weight of the object will be concentrated at the point at the distance of 5 meter from the ground and the man is at a distance of 7.5 meter from the ground.
The net force acting on the ladder should be equal to zero because the ladder is resting on the ground.
The moment of all the forces about the point where the ladder is resting we get :
S (15 sinθ) = mg ( 5 cosθ) + Mg (7.5 cosθ)
S (15) (4/5) = 500(5) 3/5 + 800(7.5) 3/5 #[g= 10]
12 S = ( 1500 + 3600 ) N
12 S = 5100 N
S = 425 N
As the net vertical force is also zero in order to be in equilibrium,
R = mg + Mg
R = ( 500 + 800 ) N
R = 1300 N
