For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force
here weight will act at mid point of door so its distance is half of the total distance where force is applied
here we know that
now we will have
so our applied force is 72.5 N
Answer:
L = 5076.5 kg m² / s
Explanation:
The angular momentum of a particle is given by
L = r xp
L = r m v sin θ
the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1
as we have two bodies
L = 2 r m v
let's find the distance from the center of mass, let's place a reference frame on one of the masses
=
i
x_{cm} =
x_{cm} =
x_{cm} =
x_{cm} = 13.1 / 2 = 6.05 m
let's calculate
L = 2 6.05 74.3 5.65
L = 5076.5 kg m² / s