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I really think that I might be overthinking it but I was thinking about a door. When you try and open it with a force, it will produce a translational AND a rotational effect on the door. Any good door you have probably doesn't translate so that means the hinges must be applying a force to oppose this translational motion (Newton's 3rd Law).

But if the hinges are applying a force, and not at the center, shouldn't it produce a torque too? Shockingly, as the force must be equal, that means the rotational torque must be equal too right?

So how can doors turn if they have equal but opposite torques applied to them? Is this similar to how objects can fall despite having no net force (air resistance balancing out gravity)

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    $\begingroup$ Torque about what axis? The hinges define the axis of rotation and the torque relative to that axis is 0 or close to it. A poorly made or installed door will likely show unnatural wear over time. $\endgroup$ – ggcg Dec 15 '18 at 11:50
  • $\begingroup$ "Any good door you have probably doesn't translate so that means the hinges must be applying a force to oppose this translational motion (Newton's 3rd Law)." Please note that this reasoning is not using Newton's third law. $\endgroup$ – Aaron Stevens Dec 15 '18 at 15:02
  • $\begingroup$ @ggcg - Torques don't have an axis, just a direction. Forces have an axis only. $\endgroup$ – ja72 Dec 15 '18 at 19:48
  • $\begingroup$ Torques is defined relative to a reference axis or point and that is usually chosen to be a fixed axis through the body or some other coordinate axis. We may be using the term differently but your comment seems incorrect. Force usually is applied to a point of contact but acts on the center of mass of the body. $\endgroup$ – ggcg Dec 15 '18 at 20:29
  • $\begingroup$ @ggcg - what you describe is the axis of rotation, which together with the line of action of the force describes the geometry of the problem. You cannot apply a torque at a point, only restrict motion about a point. In the end when you sum of torques about the center of mass, only the location of forces is important. Remember the sum of troques is $$\sum_i (\vec{\tau}_i + \vec{r}_i \times \vec{F}_i) $$ where $\vec{r}_i$ are the force application points, and $\vec{\tau}_i$ any torques applied (regardless of location). $\endgroup$ – ja72 Dec 16 '18 at 1:44
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Torque depends on the force and the distance between the hinge and the point where the force is applied.

When you pull on the door at the handle, you apply a force and there is a nonzero distance between handle and hinges, so you get torque and as a consequence rotation around the hinges.

The hinges apply their force at the rotational center, so they can't produce any torque.

There are no "equal but opposite torques" because there is only one.

Torques change rotations just as forces change translations. Ongoing rotations or translations don't require acting torques or forces.

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When you pull open a door, you are applying a force, and a torque, the cross product of force and distance is produced. However, since the door pivots and swings about the center (hinges), the hinges are unable to produce any torque since they themselves are the centre. However, they do apply a force towards the centre.

Think about it this way, there are 2 forces, but only 1 torque produced since one force is acted on the pivot point.

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To correctly address this type of problem you need to set up an appropriate set of coordinates and evaluate all torques and forces in this coordinate system. The torque is defined about an axis, or relative to some axis via the relation cross(r, F), for each force. r is the vector from the axis to the point of contact of F. For "free" bodies one can separate the motion into two components, that of the COM which is governed by the net force, and the motion about the COM which for a rigid body is pure rotation and governed by the torques. When you constrain a rigid body to be fixed at a point (spherical hinge) or along an axis (like a door) you could still describe the motion about the COM but that becomes counter intuitive. A better approach is to calculate everything relative to the fixed axis of rotation (in this case defined by the hinges). In this description there is one and only one degree of freedom needed to describe what is happening (you really have 3 degrees of rotation but the hinge constrains two of them).

The force you apply produces a torque about the axis defined by the hinge(s). There is a reaction force at the hinge (there has to be for the constraint to work). Consider a figure of the door with a hinge modeled as a cylindrical post passing through a cylindrical cuff (hole made through the door) and consider an infinitesimal gap between the post and the surface of the cuff. The hinge force is a contact force. As such there are only two possible contributions to that force (in our ideal model where the door and hinge are "rigid"). The first is a Normal force due to the contact of the two surfaces. This will point along the radial direction along a line through the center of the hinge. The second is traction between the post and the cuff, i.e. a grip tangent to the surfaces which is due to friction. The first, being normal, will never produce a torque as cross(r, F) = 0 for that force. The second will produce a torque that resists the force you are exerting to open the door. If this happens you should apply oil or WD40 to the hinge. We can assume a friction-less surface between the post and the cuff and that force goes away. In a real life situation this force should be very small or can be made arbitrarily small. Hence relative to the fixed axis of rotation the hinge produces no torque. You can also take the limit of a very small radius for the hinge and arrive at the result that the torques due to the hinge forces about the rotation axis are approximately zero.

One key to understanding the different approaches to describing the situation is that you are free to evaluate the torques and the motion in any coordinates you want.

For free objects the COM frame is ideal for describing rotation, for fixed bodies the fixed axis is ideal.

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a) The reaction force of the hinge are not equal and opposite of the applied forces on the door. They are exactly what they need to be to force the door to rotate about the hinge.

b) It is exactly the net torque about the center of mass that rotates the door. This net torque has a contribution from the applied forces and the hinge reaction.

c) It helps to construct a free body diagram and state the equations of motion before making any assumptions. Let us look at a planar simplified example:

sketch

Here a hinge at point A has unknown reaction forces $A_x$ and $A_y$. An applied force $B_y$ is applied at a point B, and the center of mass is at point C. The distance from the pivot to the COM is $c$ and the distance from the force to the COM is $d$. Let us call the swing angle $\theta$ (not shown).

  1. Kinematics - The door is hinged at A so the only allowed motion of the center of mass C is $$ \begin{aligned} \ddot{x}_C & = -c\,\dot{\theta}^2 \\ \ddot{y}_C & = c\,\ddot{\theta} \end{aligned} $$
  2. Forces - The sum of forces moves the center of mass (mass is $m$) $$ \begin{aligned} A_x & = m \ddot{x}_C = -m c \,\dot{\theta}^2 \\ A_y + B_y & = m \ddot{y}_C = m c\,\ddot{\theta} \end{aligned} $$
  3. Torques - the sum of torques about the COM rotates the body (mass moment of inertia is $I_C$) $$ \begin{aligned} d\,B_y -c A_y & = I_C \ddot{\theta} \end{aligned} $$
  4. Solution - Solve the above three equations for the pin reactions and the motion $$ \begin{aligned} A_x & = -m c \dot{\theta}^2 \\ A_y & = \left( \frac{m c (c+d)}{I_C + m c^2}-1 \right) B_y \\ \ddot{\theta} & = \left( \frac{c + d}{I_C + m c^2} \right) B_y \end{aligned} $$
  5. Explanation
    • The reaction along the x-axis only depends on the motion of the door.
    • The reaction along the y-axis is the most complex, but it becomes zero when the force is applied through the axis of percussion $d = \frac{I_C}{m c}$.
    • The rotational acceleration depends on the torque due to the applied load $(c+d)B_y$ and the mass moment of inertia about the pin $I_C + m c^2$.
  6. Effective mass - The motion of the point B of the force defines the effective mass the force sees. The acceleration along the force is $\ddot{y}_B = (c+d) \ddot{\theta}$ and thus the effective mass is $$m_{\rm effective} = \frac{B_y}{\ddot{y}_B} = \frac{I_C + m c^2}{(c+d)^2}$$

BTW - you mentioned Newton's 3rd law, which applies here on the hinge. The forces $A_x$ and $A_y$ are applied from the hinge to the door, and the equal and opposite forces are applied from the door to the hinges (and the frame or ground).

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    $\begingroup$ This is a very convoluted way to solve this problem. Your first statement is not incorrect as the COM does move. But one can choose the axis about which to analyze the torques and the standard treatment would be to place the axis at the hinge, not the COM. This makes the problem both easier to solve and easier to explain. This is a simple rigid body problem with fewer degrees of freedom when expressed in the proper variables. -1. $\endgroup$ – ggcg Dec 16 '18 at 3:20
  • $\begingroup$ @ggcg - if you express this as a 1-DOF system then you won't get an expression for the pin forces. The point of all of this is to show that by assuming the pin forces are equal and opposite of the applied force (per OP) you reach an incorrect conclusion. $\endgroup$ – ja72 Dec 16 '18 at 19:45
  • $\begingroup$ That is simply not true. $\endgroup$ – ggcg Dec 16 '18 at 19:45
  • $\begingroup$ @ggcg - Actually my answer treats the body with 1-DOF, the angle $\theta$ and the acceleration of the center of mass is strictly a function of the motion of the one DOF. So I don't understand your criticism that I am over-complicating this problem. I am using the standard method for solving rigid body mechanics. $\endgroup$ – ja72 Dec 16 '18 at 19:49
  • $\begingroup$ All of your statements about force and torque are incongruent with basic definitions. I am at a loss to understand what your are calling "standard method". Also, if a 1-dof treatment doesn't work then how are you claiming to have a 1-dof treatment that works. Can you elaborate on that? $\endgroup$ – ggcg Dec 16 '18 at 19:52

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