Horizontal Force on a Gate Hinge?

[SQUIDDO]

A door is attached to a frame by two hinges and is at equilibrium. All forces and torques are therefore balanced.

The door has a mass of 10.5kg, a width of 1.1m, and the two hinges are 0.7m apart.
Find the direction and magnitude of the horizontal force applied to the door by the top hinge?

I think I've got the vertical component of the forces sorted. The downward gravitational force on the door must be equal to the upward force provided by the hinges, with each hinge presumably supplying equal force. As for the horizontal forces, I'm at a loss.

I'd appreciate a detailed response so I can answer variations on this question in the future. Thanks for your time!

 

[Jobrag]

Think about where the centre of mass of the door is, and think why they wrote "All forces and torques are therefore balanced."

 

[SQUIDDO]

I'm afraid I've already factored the center of mass into an extended free-body diagram and am still left scratching my head.

I'm just not sure how to derive horizontal component forces from the system.
I understand that upwards force must equal downwards force:

m.g = V₁ + V₂ where V is the upwards force supplied by each hinge.

But when it comes to the horizontal aspect, I'm at a loss...

 

[Jobrag]

Think about the torque on the bottom hinge.

 

[Nickie]

As a scientist, my response would be as follows:

First, let's define the variables we have in this scenario:
- Mass of the door (m) = 10.5kg
- Width of the door (w) = 1.1m
- Distance between hinges (d) = 0.7m
- Gravitational force (Fg) = m x g (where g is the acceleration due to gravity, approximately 9.8m/s^2)

Now, let's consider the forces acting on the door at equilibrium. We have two vertical forces - the weight of the door (Fg) acting downwards, and the reaction force from the hinges (Fr) acting upwards. Since the door is at equilibrium, these two forces must be equal in magnitude.

Therefore, we can say that:
Fg = Fr

Now, let's consider the horizontal forces. Since the door is not moving horizontally, the net force in the horizontal direction must also be zero. This means that the force applied by the top hinge (Ftop) must be equal and opposite to the force applied by the bottom hinge (Fbottom).

Therefore, we can say that:
Ftop = -Fbottom

Now, let's try to find the magnitude of Ftop. To do this, we can use the concept of torque. Torque is the product of a force and the distance from the point of rotation. In this case, the point of rotation is the bottom hinge, and the distance from this point to the top hinge is (d/2).

Using the equation for torque (T = F x d), we can say that:
Ttop = Ftop x (d/2)
And for the bottom hinge:
Tbottom = Fbottom x (d/2)

Since the door is at equilibrium, the total torque must also be zero. This means that:
Ttop + Tbottom = 0
Ftop x (d/2) + Fbottom x (d/2) = 0
Ftop x (d/2) + (-Ftop) x (d/2) = 0
Ftop x (d/2) - Ftop x (d/2) = 0
0 = 0

So, we can see that our initial assumption that Ftop = -Fbottom is correct.

Now, to find the magnitude of Ftop, we can use the equation:
Ftop