Find the magnitude of the horizontal force exerted by the lower hinge

[angel120]

Homework Statement

A symmetrical door of mass 27.5 kg is 5.44 m high and 2.46 wide. The hinges are placed at the very top and very bottom of the door. Find the magnitude of the horizontal force exerted by the lower hinge.

Homework Equations

Fd=Fd
sum tau clockwise = sum tau counter-clockwise

The Attempt at a Solution

The center of mass is dead center of the door, since it's symmetrical. However, we have only done rotation for "1-dimensional" things like a table. I know that when you apply something like this to a table, you put the counterclockwise on one side, then the clockwise on the other. Since the door is vertical and has 2 points of rotation (the 2 hinges) I'm not sure how to work this out...

 

[asleight]

Homework Statement

A symmetrical door of mass 27.5 kg is 5.44 m high and 2.46 wide. The hinges are placed at the very top and very bottom of the door. Find the magnitude of the horizontal force exerted by the lower hinge.

Homework Equations

Fd=Fd
sum tau clockwise = sum tau counter-clockwise

The Attempt at a Solution

The center of mass is dead center of the door, since it's symmetrical. However, we have only done rotation for "1-dimensional" things like a table. I know that when you apply something like this to a table, you put the counterclockwise on one side, then the clockwise on the other. Since the door is vertical and has 2 points of rotation (the 2 hinges) I'm not sure how to work this out...

Gravity acts upon the center of mass of an object. So, I'm assuming the following is legit.:

$$\sum\tau=0.$$ Taking the bottom hinge as the axes of rotation, we find:
$$\tau_g=\vec{F}_g\times\vec{r}=m\vec{g}\vec{r}\sin(\theta)$$, where $$\theta=0.735 rad$$. Now, what next?

 

[thomate1]

I would approach this problem by first calculating the torque exerted by the weight of the door on the lower hinge. This can be done by finding the center of mass of the door, which we know is at the center since it is symmetrical. Then, using the equation τ = r x F, we can calculate the torque by multiplying the distance from the center of mass to the lower hinge by the weight of the door (27.5 kg x 9.8 m/s^2). This will give us the clockwise torque exerted by the weight of the door on the lower hinge.

Next, we need to consider the horizontal forces acting on the door. Since the door is not moving horizontally, we know that the sum of these forces must be zero. Therefore, we can set up an equation with the horizontal force exerted by the lower hinge as our unknown. The equation would look like this: Fh + Fh = 0, where Fh is the horizontal force exerted by the lower hinge.

Now, we need to consider the torques exerted by the horizontal forces on the door. The only horizontal force acting on the door is the one exerted by the lower hinge. We can calculate the torque exerted by this force by using the same equation as before, τ = r x F, where r is the distance from the lower hinge to the center of mass and F is the horizontal force exerted by the lower hinge.

Finally, we can set up an equation using the sum of torques being equal to zero, since the door is not rotating. This equation would look like this: τcw + τccw = 0, where τcw is the clockwise torque exerted by the weight of the door and τccw is the counterclockwise torque exerted by the horizontal force exerted by the lower hinge. We can then solve for Fh, the horizontal force exerted by the lower hinge, by setting τcw equal to -τccw (since they are equal in magnitude but opposite in direction) and solving for Fh.

In conclusion, as a scientist, I would approach this problem by first considering the torques exerted by the weight of the door and the horizontal force exerted by the lower hinge. By setting up equations based on the principles of torque and equilibrium, we can solve for the magnitude of the horizontal force exerted by the lower hinge.