Conceptual question: translational and rotational equilibrium

In summary, there are two important conditions for a rigid body to be in equilibrium: the net force on the body must be zero and the net torque must also be zero. Torque is a vector quantity and the sign is determined using the right-hand rule. In the example given, the net torque is the sum of the two individual torques, which have the same sense of rotation.
  • #1
laica
1
0

Homework Statement



There are two important conditions for a rigid body to be in equilibrium. The first condition is that the net force on a body must be zero. The second condition is that the net torque on a body must also be zero.


2,3. Relevant equations; attempt at solution.

I apologize for the simple question but I am self-studying physics and having a hard time with this concept.

So my question is: how are these two conditions not redundant? My initial reaction is that net torque must be zero if net force is zero, since [tex]\tau[/tex]=Fr. Then I drew this situation (a door revolving around a central pivot point) that seems to show net torque without net force (the 2 forces add up to zero but there is still rotation). The img is attached to this post.

The problem is, I still think net torque would be zero.
[tex]\tau[/tex]=Fr
[tex]\tau[/tex]=-Fr
[tex]\Sigma\[/tex][tex]\tau[/tex]=(Fr)+(-Fr)=0

Where am I going wrong here? Is one radius -r maybe?

Thanks in advance!
 

Attachments

  • Force.jpg
    Force.jpg
    3.6 KB · Views: 477
Physics news on Phys.org
  • #2
The torque is the vector product of the radius-vector and the force. You have to handle both F and r as vectors.

Just thinking logically, the force on the left will rotate the rod anti-clockwise, and so will the force on the right. So the torques add, while the forces cancel.

ehild
 
  • #3
You've got your directions wrong. Torque is a vector quantity. That means that when you add two or more torques together to find the net torque, you must do so with due regard to sign.

The simplest way to add torques correctly, would be to use your intuition to find the "sense" of rotation of every force vector. For instance, in your example above, the left force tries to spin the rod counter-clockwise, while the right force tries to spin it counter-clockwise as well. Therefore, the net torque is the sum of the two, since the two torques have the same sense of rotation.
Were they opposite, then the direction of rotation would be determined by which is the greater torque.A more complete approach that you'll need for further studies, when keeping track of signs is a bit more important:
How do we take care of that sign, then? We define torque as a vector perpendicular to the plane of rotation (The plane of the lever arm vector and the force vector).

And we determine the sign by way of the right-hand rule (Wiki link:http://en.wikipedia.org/wiki/Right_hand_rule)

Torque as a vector is defined as:
[tex]\vec \tau = \vec r \times \vec F[/tex]

Applying it to the situation above, taking the torque about the axis through the center of the rod, taking the positive direction as the one coming out of the plane of the page (If this is Greek to you, read up on the article on the right hand rule and vector cross product), we see a positive torque whose magnitude is [tex]2Fr[/tex].
 

1. What is translational equilibrium and how is it related to rotational equilibrium?

Translational equilibrium refers to the state in which an object's net forces in all directions are balanced, resulting in no overall change in its linear motion. Rotational equilibrium, on the other hand, refers to the state in which an object's net torque is zero, resulting in no overall change in its rotational motion. These two concepts are related because an object can only be in rotational equilibrium if it is also in translational equilibrium.

2. How do you determine if an object is in translational equilibrium?

An object is in translational equilibrium if the vector sum of all the forces acting on it is equal to zero. This means that the forces must be balanced, with equal magnitude and opposite direction. To determine if an object is in translational equilibrium, you can use the equations of Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration.

3. What is the difference between static and dynamic equilibrium?

Static equilibrium refers to the state in which an object is at rest and remains at rest, while dynamic equilibrium refers to the state in which an object is moving at a constant velocity. In both cases, the object is in translational equilibrium, but in dynamic equilibrium, the forces acting on the object are balanced at a non-zero velocity.

4. Can an object be in equilibrium if it is accelerating?

No, an object cannot be in equilibrium if it is accelerating. This is because acceleration is a change in velocity, and for an object to be in equilibrium, it must have a constant velocity. If an object is accelerating, it means that the forces acting on it are unbalanced, resulting in a change in its motion.

5. How does the center of mass affect an object's equilibrium?

The center of mass of an object is the point at which its mass is evenly distributed. In translational equilibrium, the center of mass must remain stationary, while in rotational equilibrium, the center of mass must remain on the axis of rotation. The location of the center of mass can affect an object's equilibrium, as a shift in its position can result in unbalanced forces or torque.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
394
Replies
6
Views
684
  • Introductory Physics Homework Help
Replies
17
Views
656
Replies
7
Views
237
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
267
  • Introductory Physics Homework Help
Replies
32
Views
972
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
168
  • Introductory Physics Homework Help
Replies
11
Views
999
Back
Top