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Chapter 11: Problem 4

A uniform 300 -N trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.

Short Answer

Expert verified
The required upward force is 150 N in both cases; no net force on the hinges.

Step by step solution

01

Understand the Problem

We need to calculate the net upward force and the total force exerted on a hinged door to begin opening it. We have two scenarios: (a) when force is applied at the center and (b) when force is applied at the center of the edge opposite the hinges.
02

Define the Physical System

The trapdoor can be considered as a uniform, rectangular object hinged horizontally. Its weight (300 N) acts downwards at its center of gravity, which is the center of the door. The coordinate system should be defined with the origin at the hinge line.
03

Calculate Torque for Case (a)

For case (a), the upward force is applied at the center. Torque around the hinges is needed to open the door. The force produces a torque that equals the torque due to the weight of the door:\[ \tau_{net} = F \cdot \frac{W}{2} = 300 \cdot \frac{W}{2} \]Solving yields the upward force \( F = 150 \text{ N} \).
04

Calculate Torque for Case (b)

In case (b), the force is applied at the center of the edge opposite the hinges. The lever arm is now the full width (W) of the door:\[ \tau_{net} = F \cdot W = 300 \cdot \frac{W}{2} \]Solving for \( F \) yields \( F = 150 \text{ N} \).
05

Determine Total Force on Hinges

The total force on hinges consists of two components: the supporting reaction force against the weight of the door (300 N) and the reaction to the applied force. In static equilibrium, the hinge force vertically would be zero as it balances the vertical component of the weight and applied force.

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Key Concepts

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

Understanding Torque
Torque is a fundamental concept in mechanics related to the rotational effect of a force applied at a distance from a pivot point. Imagine trying to open a door; you instinctively push near the handle rather than at the hinge to apply torque effectively. Torque (\( \tau \)) is the product of the force (\( F \)) and the lever arm distance (\( r \)) from the pivot:\[ \tau = F \times r \]In our trapdoor exercise, we're looking at how applying force at different points affects the torque generated. The door's weight creates its own torque around the hinges, and to open the door, the applied force must generate an equal or greater counter-torque. When force is applied at the geometric center, the lever arm distance is only half of the door's width, whereas it's the full width when applied at the other edge.
Achieving Equilibrium
Equilibrium is the state where all forces and torques acting on a system are balanced, meaning the system has no net movement. In the context of our exercise, equilibrium ensures the door remains stationary when the forces are balanced.
  • In static equilibrium, the sum of forces in both horizontal and vertical directions should be zero.
  • The sum of all torques around any pivot point should also be zero.
For the trapdoor, the torques created by its weight and any applied force must balance for the door to remain closed, and any additional force just beyond this balance will start opening the door.
The Role of Forces
Forces are vectors characterized by magnitude and direction, acting upon objects causing them to move, stop, or change direction. In the trapdoor problem, several forces come into play:
  • The gravitational force (weight of the door), pulling downward through its center of mass.
  • The applied force, acting to lift the door upwards.
  • Reaction forces at the hinge, which act to counterbalance these forces to maintain equilibrium.
For effective analysis, it is crucial to visualize how these forces interact. The applied force's placement determines the torque generated, impacting the door's propensity to open. Forces at play must be accurately balanced against each other to predict outcomes successfully.
Exploring Hinge Forces
Hinge forces are the reaction forces at the pivot point where the door is supported. They play a vital role in maintaining equilibrium, especially for the trapdoor in question. When the upward force needed to open the door is applied, the hinge must counteract both the downward gravitational force and any resultant horizontal force component from the applied force to keep the trapdoor balanced. The hinge force can be broken down into two components: horizontal and vertical. In our exercise:
  • The vertical hinge force must balance the weight of the door minus any vertical component of the applied force.
  • The horizontal component comes from ensuring rotational balance, preventing sideways motion as the torque is applied.
By comprehending these interactions, one can better visualize the roles the hinge forces play in determining the motion and stability of structures.

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Most popular questions from this chapter

Two people carry a heavy electric motor by placing it on a light board 2.00 \(\mathrm{m}\) long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{N},\) with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

\(\mathrm{A} 1.05-\mathrm{m}-\operatorname{lon} \mathrm{g}\) rod of negligible weight is supported at its ends by wires \(A\) and \(B\) of equal length (Fig. P11.91). The cross-sectional area of \(A\) is 2.00 \(\mathrm{mm}^{2}\) and that of \(B\) is 4.00 \(\mathrm{mm}^{2} .\) Young's modulus for wire \(A\) is \(1.80 \times 10^{11} \mathrm{Pa}\) ; that for \(B\) is \(1.20 \times 10^{11} \mathrm{Pa}\) . At what poing the rod should a weight \(w\) be suspended to produce (a) equal stresses in \(A\) and \(B\) and (b) equal strains in \(A\) and \(B ?\)

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8 \(\mathrm{N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{mm},\) what is the breaking stress of the alloy?

BIO Supporting a Broken Leg. A therapist tells a 74 -kg patient with a broken leg that he must have his leg in a cast suspended horizontally. For minimum discomfort, the leg should be supported by a vertical strap attached at the center of mass of the leg-cast system. (Fig. P11.55). In order to comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper legs (thighs) together typically account for 21.5\(\%\) of body weight and the center of mass of each thigh is 18.0 cm from the hip joint. The patient also reads that the two lower legs (including the feet) are 14.0\(\%\) of body weight, with a center of mass 69.0 \(\mathrm{cm}\) from the hip joint. The cast has a mass of \(5.50 \mathrm{kg},\) and its center of mass is 78.0 \(\mathrm{cm}\) from the hip joint. How far from the hip joint should the supporting strap be attached to the cast?

A specimen of oil having an initial volume of 600 \(\mathrm{cm}^{3}\) is subjected to a pressure increase of \(3.6 \times 10^{6} \mathrm{Pa}\) , and the volume is found to decrease by 0.45 \(\mathrm{cm}^{3} .\) What is the bulk modulus of the material? The compressibility?
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