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Chapter 12: Problem 86

A trap door in a ceiling is \(0.91 \mathrm{~m}\) square, has a mass of \(11 \mathrm{~kg} .\) and is hinged along one side, with a catch at the opposite side. If the center of gravity of the door is \(10 \mathrm{~cm}\) toward the hinged side from the door's center, what are the magnitudes of the forces exerted by the door on (a) the catch and (b) the hinge?

Short Answer

Expert verified
(a) Force on the catch: 42.30 N, (b) Force on the hinge: 65.81 N.

Step by step solution

01

Determine the center of mass from the hinge

The trap door is a square with side length of \(0.91 \text{ m}\), and the center of gravity is \(0.10 \text{ m}\) from the center of the door toward the hinge. Therefore, the distance from the hinge to the center of mass is \((0.91/2 - 0.10) = 0.355 \text{ m}\).
02

Calculate the torque about the hinge

The torque \(\tau\) due to the weight of the door can be calculated using the formula \(\tau = r \cdot F \cdot \sin(\theta)\), where \(r\) is the distance from the hinge to the center of gravity \(0.355 \text{ m}\), \(F = 11 \cdot 9.81 \text{ N}\) is the force due to gravity, and \(\theta = 90^\circ\) because gravity acts vertically. So, \(\tau = 0.355 \times 11 \times 9.81 = 38.48 \text{ Nm}\).
03

Force at the catch due to torque

The torque caused by the door's weight needs to be balanced by the force at the catch since the door is hinged. The distance from the hinge to the catch is \(0.91 \text{ m}\). Let \(F_{catch}\) be the force at the catch, then \(F_{catch} \times 0.91 = 38.48 \). Solving for \(F_{catch}\), \(F_{catch} = \frac{38.48}{0.91} = 42.30 \text{ N}\).
04

Calculate vertical forces at the hinge

The sum of vertical forces must equal the weight of the door. Therefore, if \(F_{hinge}\) is the vertical component of the hinge force, then \(F_{catch} + F_{hinge} = Weight\). Hence, \(F_{hinge} = 11 \times 9.81 - 42.30 \). This gives \(F_{hinge} = 108.11 - 42.30 = 65.81 \text{ N}\).

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Key Concepts

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

Torque calculation
Torque plays a crucial role in understanding rotational forces. It refers to the rotational equivalent of linear force and determines how effectively a force can cause an object to rotate around a pivot.
To calculate torque, we use the formula:
\[ \tau = r \cdot F \cdot \sin(\theta) \]where:
  • \(\tau\) is the torque.
  • \(r\) is the distance from the pivot or hinge to the point where the force acts.
  • \(F\) is the magnitude of the force.
  • \(\theta\) is the angle between the force and the lever arm, typically 90 degrees in vertical setups.
In our trap door problem, the force due to gravity applies vertically, so \(\sin(90^\circ) = 1\). Therefore, the torque becomes a simple product of the force and distance from the hinge to the center of mass.
Center of mass
The concept of the center of mass is essential in physics as it defines the average location of an object's entire mass. For symmetric objects like our square trap door, the center of mass generally lies along the geometric center, but modifications can shift it slightly.
The trap door's center of mass is given 10 cm toward the hinge from its center. This adjustment in position can affect calculations of torque and balance.
By determining this shifted center, we can accurately calculate the distance from the hinge that impacts the torque. For the trap door, this calculated distance is 0.355 meters from the hinge, factoring directly into torque computations.
Equilibrium of forces
Equilibrium is when an object is in a state where all forces are balanced, and there's no net motion. It鈥檚 a key aspect when working with structures like our trap door.
For an object like a door that's hinged on one side, equilibrium demands that rotational forces (torques) and linear forces must balance.
  • The torque due to the weight must be counteracted by the force at the catch since no net rotation occurs.
  • All vertical forces acting on the door鈥攂eing at the catch and hinge鈥攎ust together equal the door's weight.
When these conditions of equilibrium are satisfied, the system remains stable without any unplanned movement or rotation. This ensures safety and functionality in practical applications like ceiling trap doors.
Vertical forces
Vertical forces often significantly influence an object, particularly in gravitational contexts. In our exercise, the vertical forces include the weight of the door and the reactions at the catch and hinge.
The total vertical force exerted by these components must equal the gravitational force on the door, ensuring the trap door hangs in a stable position without collapsing under its weight.
Here's how they balance:
  • The weight provides a downward force calculated using \( F = m \cdot g \), where \(m\) is mass and \(g\) is acceleration due to gravity.
  • The vertical force at the catch and hinge must sum to this weight to maintain equilibrium.
In our example, analyzing these forces allowed us to solve for unknown forces at both the catch and hinge.

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Most popular questions from this chapter

A gymnast with mass \(46.0 \mathrm{~kg}\) stands on the end of a uniform balance beam as shown in Fig. \(12-80\). The beam is \(5.00 \mathrm{~m}\) long and has a mass of \(250 \mathrm{~kg}\) (excluding the mass of the two supports). Each support is \(0.540 \mathrm{~m}\) from its end of the beam. In unit-vector notation, what are the forces on the beam due to (a) support 1 and (b) support \(2 ?\)

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