/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A door \(1.00 \mathrm{~m}\) wide... [FREE SOLUTION] | 91Ó°ÊÓ

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A door \(1.00 \mathrm{~m}\) wide and \(2.00 \mathrm{~m}\) high weighs \(330 \mathrm{~N}\) and is supported by two hinges, one \(0.50 \mathrm{~m}\) from the top and the other \(0.50 \mathrm{~m}\) from the bottom. Each hinge supports half the total weight of the door. Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

Short Answer

Expert verified
The horizontal components of force exerted on the door by each hinge are both \(220N\).

Step by step solution

01

Identify Forces

Firstly, identify all the forces acting on the door. The weight of the door acts downward from its center. There are also vertical and horizontal forces at each hinge. Let the horizontal forces at the top and bottom hinge be \(F_{T}\) and \(F_{B}\) respectively.
02

Apply Equilibrium Conditions

The door is in equilibrium. Therefore, sum of the torques about any point is zero. We can choose the bottom hinge as our point of reference, as this will simplify the calculations. The horizontal forces exert no torque as they pass through the point of consideration, while the weight of the door and the vertical force of the top hinge create torques of equal magnitude but opposite directions.
03

Calculate the Torques

The torque due to the weight is its force (\(330N\)) times the distance from the point of reference, which is \(1.00m\). The torque due to the vertical force at the top hinge is half the weight (\(165N\)) times the distance from the point of reference, which is \(1.50m\). Setting these two torques equal gives: \(330N \cdot 1.00m = 165N \cdot 1.50m\).
04

Solve for Force

Solving the equation gives: \(F_{T} = F_{B} = \frac{330N \cdot 1.00m}{1.50m} = 220N\). Therefore, the horizontal components of force exerted on the door by each hinge is \(220N\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque Calculation
Understanding torque is crucial when analyzing forces and motion in mechanics. Torque, often referred to as the 'rotational equivalent of force', is the measure of the force causing an object to rotate about an axis. The calculation for torque (\tau) involves two key components: the magnitude of the force (F) and the perpendicular distance from the axis of rotation to the line of action of the force (r), expressed in the equation: \[ \tau = r \times F \]
In the case of the door mentioned in our exercise, the torque calculation involves both the weight of the door and the vertical forces at the hinges. By understanding that the distance from the point of reference to the force's line of action is a lever arm, students can comprehend how applying force at different points results in varying torques. Remember to multiply the force by the perpendicular distance from the pivot point for an accurate calculation of the torque.
Equilibrium Conditions
When studying forces in mechanics, the concept of equilibrium is essential. An object is in equilibrium when all the forces and torques acting upon it are balanced, meaning there is no unbalanced force or torque to cause linear or rotational acceleration respectively. For objects in static equilibrium, two conditions must be met:
  • The sum of all horizontal forces must be zero.
  • The sum of all torques about any point must be zero.

For our example of the door, we utilize the equilibrium conditions to determine the horizontal forces at the hinges. By selecting the most convenient pivot point, which often means one that will eliminate some forces from the torque calculation, we simplify our task. This approach is visible in the solution where the bottom hinge is chosen, negating the need to account for its horizontal force, thus only requiring the calculation of the torque generated by the door's weight and top hinge force.
Forces in Mechanics
Forces are essential to understand when exploring mechanics, as they are influencers of motion. Every force has both magnitude and direction and can cause an object to accelerate, slow down, or change direction. In the context of our door example, the forces include the weight of the door and the forces at the hinges.

Breaking Down Forces:

The weight is a downwards force due to gravity, acting at the door's center of gravity. The hinges provide the reaction forces to the door's weight, and these are typically broken into vertical and horizontal components. The vertical components support the door's weight, while the horizontal components, which we are asked to find in the problem, prevent the door from rotating about the vertical axis. Students should visualize the multidirectional aspect of forces and understand that the forces in mechanics are vectors, hence they obey vector addition. Understanding all these components helps in analyzing the equilibrium of objects and solving related problems.

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Most popular questions from this chapter

In lab tests on a \(9.25 \mathrm{~cm}\) cube of a certain material, a force of \(1375 \mathrm{~N}\) directed at \(8.50^{\circ}\) to the cube (Fig. E11.38) causes the cube to deform through an angle of \(1.24^{\circ} .\) What is the shear modulus of the material?

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A metal rod that is \(4.00 \mathrm{~m}\) long and \(0.50 \mathrm{~cm}^{2}\) in crosssectional area is found to stretch \(0.20 \mathrm{~cm}\) under a tension of \(5000 \mathrm{~N}\). What is Young's modulus for this metal?

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