Equilibrium Rotation Physics. Hinged Door Please help?

Question

A door has a height of 2.1 m along a y axis that extends vertically upward and a width of 0.99 m along an x axis that extends outward from the hinged edge of the door. A hinge 0.30 m from the top a hinge 0.30 m from the bottom each support half the door's mass which is 23 kg.

What is force on door at top hinge?
(N i + Nj)???

What is force on door at bottom hinge?
(Ni + Nj)

Answer 1

The weight of this door is 23 × 9.8 = 225.4 N. Its center of mass is located just where you'd expect it to be, at the point where the diagonals cut each other. To be more specific, If the lower left corner of the door is at the origin, the center of mass has coordinates (0.495, 1.05).

Only the x-coordinate is relevant here, because it is the lever-arm over which door's weight operates to produce torque. Each hinge supports half of the weight, so the vertical component of the force, at each hinge, is 112.7 N, upwards.

There are also horizontal components of force at each hinge. These x-components constitute what is called "a couple". They are exact opposites, so no net force is exerted on door. However, these forces aren't aligned with a common "line-of-action", and, therefore, they can -and do- originate a torque.

This torque counteracts the torque the door's weight exerts on it. You can choose whatever point you please to act as pivot, but torques may be easier to figure out if you choose any hinge as pivot. Placing the pivot point at the lower hinge, you get

1.5 Fx + 0.495 × 225.4 = 0

where Fx is the horizontal force the upper hinge exerts on door. Net torque should be zero, since the door doesn't turn about any point. Solve the above equation for Fx to get Fx = -74.38 N. Fx points in the negative direction of the x-axis, in order to counteract torque due to weight. At the lower hinge, an equal, but opposite, force acts as well on the door.

At the top hinge, the force is thus
(-74.38i + 112.7j ) N.

At the bottom hinge,
(74.38i + 112.7j) N.

Answer 2

The distance from a hinge to the center of mass of he door is R

The distance between hinges is 2.1-.6=1.5

The angle subtended by R to the center of mass and the vertical edge of the door is th, where sin(th)=.495/R

Summing torques about the bottom hinge,
sin(th)*m*g*R=Tx*1.5
Tx=.495*23*9.81
Tx=111.7 N

The top hinge will pull the door in the -x direction
The bottom hinge will push the door in the x direction, the force magnitudes will be equal.

The vertical forces are evenly distributing the vertical weight of the door between the two hinges, Ty=23*9.81/2
Ty=23*9.81/2
Ty=112.815 N

The total force magnitudes are
Sqrt(112.815^2+111.7^2)

=158.8 N

j

Answer 3

"gentle My fireplace" it incredibly is unhappy. we are dropping an excellent number of great ones from that era. to boot to being a proficient and proficient musician and singer (additionally producer, action picture director, and author), Ray Manzarek gave the impression of an extremely effective guy, smart, articulate, down-to-earth, and with a great concepts-set in the direction of life. i'm specific he's jamming with Jim Morrison. There ought to be some marvelous track in Heaven.