Torque: Turning Heavy Door with "Steady" Force

In summary, the question asks about the required steady force to move a heavy hinged door through an angle of 90 degrees in 30 seconds, neglecting friction. The correct solution involves using the equation for angular velocity and acceleration, as well as the torque equation, and taking into account the mass of the door. The mistake made in the attempted solution was not multiplying by pi in one of the equations.
  • #1
JFlash
6
0
[SOLVED] Torque: Turning Heavy Door with "Steady" Force

Homework Statement



Hello. The question reads, "...This is the world's heaviest hinged door. The door has a mass of 44,000kg, a rotational inertia about an axis through its hinges of 8.7X10^4 kg*m^2, and a width of 2.4m. Neglecting friction, what steady force, applied at its outer edge and perpendicular to the plane of the door, can move it from rest through an angle of 90 degrees in 30s? Assume no friction acts on the hinges."

Homework Equations



The work I did below followed 4 steps:
1. Use this equation: [tex]\Theta = (1/2)(\omega initial + \omega)t[/tex] to find the angular velocity at 30s.
2. Find the average angular acceleration over the 30s period of applied force.
3. Set the torque equal to the rotational inertia of the door times its angular acceleration.
4. Set torque equal to r X F, and solve for F.

The Attempt at a Solution



My answer came to 40.27N, which is about a third of the actual answer: 130N. As I look over my work, I realize that I never used the mass of the door, which is probably a reason why my answer is wrong. Anyway, here's what I did:

[tex]\Theta = \pi/2 = (1/2)(0 + \omega)30s... \omega = \pi/30 rad/s[/tex]
[tex]\alpha = (\pi/30 rad/s)/(30s-0s) = \pi/900 rad/s^2 [/tex]
[tex]\tau = 87000kgm^2 * \pi/900 rad/s^2 = 96.67[/tex]
[tex]\tau = rF sin \phi = 2.4m * F * sin(90 degrees) = 40.27N[/tex]

Does anyone know where I went wrong?
 
Last edited:
Physics news on Phys.org
  • #2
you forgot to multipy by pi in: [tex]\tau = 87000kgm^2 * \pi/900 rad/s^2 = 96.67[/tex]
 
  • #3
Thanks man.
 

Related to Torque: Turning Heavy Door with "Steady" Force

1. What is torque and how is it related to turning a heavy door?

Torque is a measure of the rotational force applied to an object. In the case of turning a heavy door, torque is the force that is applied to the door handle in order to rotate the door around its hinges.

2. How is torque different from regular force?

While force is a measure of the push or pull applied to an object, torque is specifically a measure of the force that causes an object to rotate around an axis. In other words, torque is a rotational force, while regular force can be applied in any direction.

3. What factors affect the amount of torque needed to turn a heavy door?

The amount of torque needed to turn a heavy door depends on several factors, including the weight of the door, the distance from the hinge to the handle, and the friction between the door and its hinges. The farther the handle is from the hinge and the heavier the door, the more torque will be required. Additionally, more friction between the door and its hinges will also require more torque.

4. Can torque be increased or decreased?

Yes, torque can be increased or decreased by changing the amount of force applied or by changing the distance from the axis of rotation. For example, using a longer handle to open a door will require less force to achieve the same amount of torque compared to using a shorter handle.

5. How is torque measured?

Torque is typically measured in units of newton-meters (N⋅m) or foot-pounds (ft⋅lb) in the metric and imperial systems, respectively. It can be calculated by multiplying the amount of applied force by the distance from the axis of rotation to the point where the force is applied.

Similar threads

  • Introductory Physics Homework Help
Replies
32
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
942
  • Introductory Physics Homework Help
Replies
3
Views
11K
Replies
52
Views
4K
  • Introductory Physics Homework Help
3
Replies
88
Views
21K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top