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A uniform \(300 \mathrm{~N}\) 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 net upward force needed to begin to open the door and the total force exerted by the hinges: a) if the upward force is applied at the center is 300N each. b) If the upward force applied at the center of the edge opposite the hinges is 150N and the total force at the hinge would be 450N.

Step by step solution

01

Identifying Given Parameters

We have a trapdoor of weight 300N, which acts at the center of gravity (midway between the hinge and the opposite edge). The weight can be considered as a force exerted on the door at its center.
02

The upward force is applied at the center

The pivot point is the hinge. If the upward force is applied at the center, it is equidistant from the hinge and thus results in no net torque due to symmetry. This means that the net upward force required to open the door is equal in magnitude to the weight of the door i.e., 300N. The total force exerted on the door by hinges is also equal to 300N to balance the weight of the door and the applied force.
03

The upward force applied at the center of the edge opposite the hinges

The pivot point is still the hinge, but the force applied is twice as far from the hinge as the weight. To produce just enough torque to initiate rotation (open the trapdoor), the applied force is half the trapdoor’s weight = \(300 N/2 = 150 N\). The total force exerted by the hinges would be the vector sum of this applied force and the weight of the door i.e., 300 + 150 = 450 N
04

Conclusion

The location of the force applied to open the trapdoor changes the amount of force needed as well as the total force exerted by the hinges to maintain equilibrium. Applying the force at the center of the door requires more force but results in less overall force on the hinges.

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Key Concepts

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

Torque
Torque is an essential concept in the study of statics, especially when analyzing rotating systems or objects about a pivot point. It is the measure of the force that can cause an object to rotate about an axis. The magnitude of torque depends on three main factors:
  • The amount of force applied
  • The distance from the pivot point to where the force is applied
  • The angle at which the force is applied
Torque is calculated using the formula: \[\tau = r \cdot F \cdot \sin(\theta) \] Where \(\tau\) is the torque, \(r\) is the distance from the pivot point, \(F\) is the magnitude of the force, and \(\theta\) is the angle between the force vector and the lever arm.
In the exercise about the trapped door, we analyze how the distribution of force at different points affects torque. When force is applied at the center where the distance \(r\) is smaller, more force is needed because the lever arm length is shorter. Conversely, when the force is applied at the edge, the lever arm is longer, thus needing less force to produce the same torque necessary to initiate rotation.
Equilibrium
Equilibrium in physics means that all the forces acting on a system are balanced, which results in no net movement or rotation. In static equilibrium, the sum of all forces and torques around any given axis is zero. This is represented by the following conditions:
  • Sum of all vertical forces is zero
  • Sum of all horizontal forces is zero
  • Sum of all torques around a pivot is zero
In the trapdoor exercise, equilibrium is vital to determine the forces needed to maintain balance when an upward force is applied. When the force is applied at the center, equilibrium must be met as both sides have equal torque, hence the total force required is equal to the door's weight to maintain balance.
If the force is applied farther out, achieving balance requires adjusting the magnitude of the applied force, according to its distance from the hinges. This ensures the door's weight and applied force maintain rotational balance.
Forces
Forces are interactions that change the motion of an object. They can be caused by various factors such as gravity, friction, or applied forces. Forces are quantified in newtons (N) and can either be contact forces (such as friction) or action-at-a-distance forces (such as gravity).
In the context of the trapdoor problem, understanding forces is crucial. The door has a weight of 300N due to gravity. This weight acts as a downward force at its center of gravity. When examining the steps to open the door, a force is applied upwardly to counteract this weight.
  • An upward force equal to the weight (300N) is needed at the center to begin opening the trapdoor since it perfectly opposes the gravitational pull.
  • When the force is moved to the edge, only 150N is required because the leverage provided by the increased distance results in a greater torque.
These principles show how the concepts of forces, weight, and their application points influence the mechanical advantage needed in practical applications. Understanding these fundamentals is key to solving complex static problems.

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Most popular questions from this chapter

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?

A \(60.0 \mathrm{~cm}\), uniform, \(50.0 \mathrm{~N}\) shelf is supported horizontally by two vertical wires attached to the sloping ceiling (Fig. E11.10). A very small \(25.0 \mathrm{~N}\) tool is placed on the shelf midway between the points where the wires are attached to it. Find the tension in each wire. Begin by making a free-body diagram of the shelf.

A uniform rod is \(2.00 \mathrm{~m}\) long and has mass \(1.80 \mathrm{~kg} .\) A \(2.40 \mathrm{~kg}\) clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be \(1.20 \mathrm{~m}\) from the left-hand end of the rod?

You are a construction engineer working on the interior design of a retail store in a mall. A 2.00 -m-long uniform bar of mass \(8.50 \mathrm{~kg}\) is to be attached at one end to a wall, by means of a hinge that allows the bar to rotate freely with very little friction. The bar will be held in a horizontal position by a light cable from a point on the bar (a distance \(x\) from the hinge) to a point on the wall above the hinge. The cable makes an angle \(\theta\) with the bar. The architect has proposed four possible ways to connect the cable and asked you to assess them: $$ \begin{array}{lllll} \text { Alternative } & \text { A } & \text { B } & \text { C } & \text { D } \\\ \hline x(\mathrm{~m}) & 2.00 & 1.50 & 0.75 & 0.50 \\ \theta(\text { degrees }) & 30 & 60 & 37 & 75 \end{array} $$ (a) There is concern about the strength of the cable that will be required. Which set of \(x\) and \(\theta\) values in the table produces the smallest tension in the cable? The greatest? (b) There is concern about the breaking strength of the sheetrock wall where the hinge will be attached. Which set of \(x\) and \(\theta\) values produces the smallest horizontal component of the force the bar exerts on the hinge? The largest? (c) There is also concern about the required strength of the hinge and the strength of its attachment to the wall. Which set of \(x\) and \(\theta\) values produces the smallest magnitude of the vertical component of the force the bar exerts on the hinge? The largest? (Hint: Does the direction of the vertical component of the force the hinge exerts on the bar depend on where along the bar the cable is attached?) (d) Is one of the alternatives given in the table preferable? Should any of the alternatives be avoided? Discuss.

The left-hand end of a uniform rod of mass \(2.00 \mathrm{~kg}\) andlength \(1.20 \mathrm{~m}\) is attached to a vertical wall by a frictionless hinge. The rod is held in a horizontal position by an aluminum wire that runs between the right-hand end of the rod and a point on the wall that is above the hinge. The cross-sectional radius of the wire is \(2.50 \mathrm{~mm}\) and the wire makes an angle of \(30.0^{\circ}\) with the rod. (a) What is the length of the wire? (b) An object of mass \(90.0 \mathrm{~kg}\) is suspended from the right-hand end of the rod. What is the increase in the length of the wire when this object is added? In your analysis do you need to be concerned that the lengthening of the wire means that the rod is no longer horizontal?

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