Torque- rigid objects in equalibrium/ center of gravity.

[utnip123]

Homework Statement

A uniform door(.81m wide and 2.1m high) weighs 140N and is hung on two hinges that fasten the long left side of the door to a vertical wall. The hinges are 2.1m apart. Assume that the lower hinge bears all the weight of the door. Find the magnitude and direction of the horizontal component of the force applied to the door by a) the upper hinge and the lower hinge. Determine the magnitude and direction of the force applied by the door to b) the upper hinge and the lower hinge.

Fg = 140 N

Homework Equations

ƩT = F L
am stuck on how the torque equation applies to this problem

The Attempt at a Solution

i drew the diagram of the door and the hinges spaced 2.01 m apart.
I think you have to make the pivot point/axle point at the bottom hinge

sum of the verticle forces = weight of the door, I am assuming this.
Am not sure how to even begin this question. but do i start to make x components at the hinges?

any advice would be helpfull.

thanks

 

[SammyS]

Homework Statement

A uniform door(.81m wide and 2.1m high) weighs 140N and is hung on two hinges that fasten the long left side of the door to a vertical wall. The hinges are 2.1m apart. Assume that the lower hinge bears all the weight of the door. Find the magnitude and direction of the horizontal component of the force applied to the door by a) the upper hinge and the lower hinge. Determine the magnitude and direction of the force applied by the door to b) the upper hinge and the lower hinge.

Fg = 140 N

Homework Equations

ƩT = F L
am stuck on how the torque equation applies to this problem

The Attempt at a Solution

i drew the diagram of the door and the hinges spaced 2.01 m apart.
I think you have to make the pivot point/axle point at the bottom hinge

sum of the verticle forces = weight of the door, I am assuming this.
Am not sure how to even begin this question. but do i start to make x components at the hinges?

any advice would be helpfull.

thanks

Hello utnip. Welcome to PF !

Assume that gravity acts downward on the door at its center. That will produce a torque about the hinges. Find the torque about the lower hinge.

 

[briantaekim]

See image below

 

[vizor-alate]

but where is your axis on the door?

 

[Nickie]

I would approach this problem by first defining the concepts involved. Torque is a measure of the rotational force applied to an object. In this case, the door is a rigid object and is in equilibrium, meaning that all forces acting on it are balanced. The center of gravity is the point where the weight of the object can be considered to act. In this problem, the center of gravity of the door is located at its midpoint, 1.05m from each hinge.

To solve this problem, we can use the equation ƩT = F x L, where ƩT is the sum of the torque, F is the force applied, and L is the lever arm (the perpendicular distance from the pivot point to the line of action of the force). In this case, the pivot point is at the lower hinge, so the lever arm for the upper hinge is 2.1m and the lever arm for the lower hinge is 0m (since the force is acting at the pivot point).

To find the magnitude and direction of the horizontal component of the force applied by the hinges, we can set up the following equations:

ƩT = 0 (since the door is in equilibrium)
Tupper + Tlower = 0 (since the door is not rotating)
(Tupper)(2.1m) + (Tlower)(0m) = 0 (since the lever arm for the lower hinge is 0m)

Solving for Tupper, we get Tupper = -Tlower. This means that the magnitude of the horizontal component of the force applied by the upper hinge is equal to the magnitude of the horizontal component of the force applied by the lower hinge, but in the opposite direction. This makes sense since the door is in equilibrium, so the forces must be balanced.

To find the magnitude and direction of the force applied by the door to the hinges, we can set up the following equations:

ƩT = Fg x L (since the door is not rotating)
Tupper + Tlower = Fg x L (since the door is not rotating)
(Tupper)(2.1m) + (Tlower)(0m) = (140N)(1.05m) (since the lever arm for the lower hinge is 0m and the lever arm for the upper hinge is 2.1m)

Solving for Tupper and Tlower, we get Tupper =