(III) A door 2.30 m high and 1.30 m wide has a mass of 13.0...

Name: Problem 29, Chapter 9: Physics: Principles with Applications
Uploaded: 2019-09-05T23:33:58-08:00
Duration: 6 min 1 s
Channel: Kebur Fantahun
Description: (III) A door 2.30 m high and 1.30 m wide has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door's weight (Fig. 9-69). Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door. FIGURE 9–69 Problem 29. (Figure can't copy)

(III) A door 2.30 m high and 1.30 m wide has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door's weight (Fig. 9-69). Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door. FIGURE 9–69 Problem 29. (Figure can't copy)

Key Concepts

Static Equilibrium

This principle states that for a body to be at rest, the sum of all forces and the sum of all torques (moments) acting on it must be zero. In problems like this door analysis, static equilibrium is used to set up equations that balance both the translational (force) and rotational (torque) effects to solve for unknown forces at the hinges.

Center of Mass

The center of mass is the point in an object where the entire mass can be considered to be concentrated. In uniformly distributed mass problems, like a door with a uniform thickness, the center of gravity is located at the geometrical center. This concept is essential for determining how the weight of the door contributes to the overall torque about any point.

Torque (Moment) Calculation

Torque is the measure of the force that causes an object to rotate about an axis and is calculated as the product of the force and the perpendicular distance from the pivot point (lever arm). In problems like this, determining the torques about a hinge helps in setting up the rotational equilibrium equations necessary to find the unknown reaction forces.

Force Decomposition into Components

Force decomposition involves breaking a force into its horizontal and vertical components. In hinge problems, it is necessary to resolve the reaction forces into these orthogonal components to satisfy the equilibrium conditions in both the horizontal and vertical directions, ensuring a complete analysis of the forces acting on the door.

Transcript

00:01 Chapter 9, problem 29.

00:02 A door, 2 .3 meters high and 1 .3 meters wide, has a massive 13 kilograms.

00:06 A hinge, 0 .4 meters from the top and another one from the bottom.

00:12 Both support half the door's weight.

00:15 And what this tells us is basically that the wide component of the force from both of the hinges is equal and it's equal to half the weight.

00:23 So this is what we see on the first page.

00:27 It's the second statement in the question.

00:30 At the center of gravity is at the geometrical center of the door and determine the horizontal and vertical force components exerted by each inch on the door.

00:40 So here's our free bottom diagram of the door, or here's our diagram of the door, i should say.

00:48 The width is here, denoted by w, height by h, center of mass by the dot in the middle with a little m .g.

00:54 There's a distance d if you notice here and here as well.

01:00 So 0 .4 from the bottom and point four from the top is where the hinges are.

01:05 And that's where we have these little local force areas is at the hinges.

01:11 So a is the top hinge and b is the bottom hinge.

01:17 So like i said before, the forces on the door are due not only to the hinges, but to the door's weight too, but the fact of the matter is that both white components are equal to half the weight, and we'll use that later.

01:32 Another important fact is that we have the door not moving, no acceleration, so this means that we can use some of the forces and the sum of the torques equal to 0, equal to ma, but in 0 in this case, it's not moving, to find some stuff.

01:47 So the door is in equilibrium, so the sum of the torques is equal to some of the forces is equal to 0.

01:55 One thing that we want to focus on is to treat the bottom hinge as the pivot.

02:00 So if we go back to our picture here, our diagram, right? so if we treat the bottom hinge here as the pivot, let me do blue for the hinge.

02:12 Right, so yeah.

02:17 So if we treat this as the pivot point, then we essentially don't need to worry about that force anymore.

02:24 But we still have torques being created by the other forces, right? that we can say that counterclockwise is going to be positive...

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