# What rotates a door, torque or force?

I am confused as to what causes a door to rotate. Assume I apply a force to the edge of the door, with the door initially at rest. Force is perpendicular to the door, so it would create a torque. But isn't there also a force on the door? The hinge itself sees very little force, so isn't the door getting a force on the centre of mass and a torque? Why doesn't the door then rotate AND push out of the frame?

Take, for example, a sphere in space. If you apply a force, it will translate according to $$F=ma$$, but it will rotate according to $$\tau=I\alpha$$.

So for the door, doesn't the door have both a torque and force on it? Which is the cause of its rotation, and how does a net force on the door work then?

• The key here is: "causes a door to rotate". Rotation is caused by torque: $\tau=I\alpha$ where $\alpha$ is the angular acceleration. The door doesn't also translate because the hinge counteracts the torque-causing force.
– Gert
Mar 1, 2021 at 19:43
• But the Center of Mass does translate. I agree the hinge takes a force, and it is balanced by whatever it is attached too. However, the door still has a net force about its center of Mass at the moment it starts to rotate Mar 1, 2021 at 20:09
• But the CoM of the door moves also, no? If you think of this in a coordinate system attached to the CoM there is both a torque rotating the door about its CoM, and a force moving the CoM. The two effects "miraculously" combine resulting in a motion where the side of the door constrained by the hinges only moves very little. Mar 1, 2021 at 20:28
• @Jyrki Lahtonen, so you are saying the Cof mass moved but it rotates, so it is both force and torque moving the door? Mar 1, 2021 at 20:34
• @Gert The door does translate, since its center of mass is not stationary. The hinges are directly responsible for this trajectory - you can push a door directly away from you, and its CoM will move away and then to the right or left as it swings open. You never provided any leftward/rightward force at all, yet the door's CoM translates right or left. That left/right force comes from the hinges. A revolving door doesn't translate (the CoM never moves), but a swinging door does. Mar 1, 2021 at 20:39

There are three forces on the door: gravity $$\vec F_g$$, the applied force $$\vec F_a$$, and the constraining force from the door hinges $$\vec F_h$$. $$\vec F_g + \vec F_a + \vec F_h = m\vec a$$ where $$a$$ is the acceleration of the center of mass (CM) of the door and $$m$$ is the mass of the door. The component of the hinge force in the vertical direction counters both the force of gravity and any component of the applied force downward, if there is any. The hinges provide a constraint causing rotational motion about a fixed axis vertical up through the hinges. Taking the torque about this axis $$\vec \tau = \vec F_a \times \vec d = I \vec {\dot\omega}$$ where $$I$$ is the moment of inertia of the door about the axis of rotation, $$\vec \omega$$ is the angular velocity of the door about the axis, and $$\vec d$$ is the vector from the axis of rotation to the point on the door where $$\vec F_a$$ is applied. The hinge force does not have a torque about this axis of rotation since the moment arm for the hinge force is zero, and gravity does not provide a torque about this axis. The CM motion is based on the net force, rotation of the door about the fixed axis of rotation is based on the net torque. Both forces and torques contribute to the overall motion of the door. The net force is greater than zero for the CM to move. The net torque about the axis of rotation is greater than zero for the door to rotate about the axis. Both a net force and a net torque cause the motion.

I suggest you draw out a free body diagram showing the forces and torques; use polar coordinates about the axis of rotation. For simplicity, assume $$\vec F_a$$ has no vertical component and is always perpendicular (90 degrees) to $$\vec d$$.

• so becuase we have both rotational and translation motion, there is both net torque and force correct? Also, assuming I had an inside door of a house, not moving, then someone pushes it. the force applied minus the force from the hinge would equal MA for the motion of the c of mass correct? Want to make sure I understand your explanation. Mar 1, 2021 at 22:51
• Yes, but the vertical components of all the forces sum to zero since the CM does not move vertically. Mar 1, 2021 at 22:57
• I agree with that. I'm ignoring vertical. From what I understand, really the force induces a torque and due to the hinge, the forces about the c of mass must balance to follow the motion. Many times we are only concerned about the rotation of the door, not necessarily the c of mass motion, which makes sense why many textbooks leave this detail out. This also agrees with the example above. Thank you Mar 1, 2021 at 23:05
• Yes, they really should also at least mention the motion of the CM to show that although the hinge force does not contribute to the torque it affects the motion of the CM, as the example in your question shows (force of hinge on rod = - force of rod on hinge). Mar 1, 2021 at 23:33

Consider a door floating in outer space, not attached to anything. If you apply a force to the door at its center of mass and perpendicular to the door, it will be displaced in the direction of the force with an acceleration given by Newton's 2nd Law, but it will not rotate. If, on the other hand, you apply a force perpendicular to the door and at the edge of the door, it will rotate AND be displaced.

The difference between the above scenario and a door in your house is that the door in your house is on a hinge. When you push on the edge of the door opposite the hinge, the door only rotates because the hinge is applying enough force to keep the door from translating. Even though that force is "small", it still exists.

This all means that any force that is not through the center of mass of an object such as a door will cause a torque that will rotate that object unless there is another torque that stops that rotation from happening. That is the "normal" case, and the case of a force causing only translation is a special case where the summation of all torques leads to zero net torque.

• but how do you explain the c of mass of the door initially does move, so there is a force on the door along with torque. The hinge force is there, i agree, but it does not equal the input force. So it isnt a pure torque Mar 1, 2021 at 20:37
• I don't understand what your concern is. What does "The hinge force is there, i agree, but it does not equal the input force" mean? What is the "input force" and why do you think that the hinge force must equal it? Also, how do you know what the hinge force is without doing a detailed free body diagram? Mar 1, 2021 at 20:40
• i have edited my question. See above. IF you look, there is a net force and net torque. So back to my question, it is the torque or force responsible for making the door rotate Mar 1, 2021 at 20:50
• When you push on the door, you exert both a torque and a force. The motion is due to the combination of all forces / torques - you can't really say "the force does this, the torque does that". Also note that a torque with respect to an axis that does not pass through the center of mass of the object can cause the center of mass to accelerate Mar 1, 2021 at 21:26
• @JakobKS, you should have made that the answer. That makes the most sense! So the torque cuases there to be acceleration of the center of mass since it is off axis. So in a sense, the torque causes a rotation and the forces about the hinge/c of mass must follow to keep everything balanced! Got it . Thank you Mar 1, 2021 at 21:37

It is possible to analyze the total movement as a sum of rotation of the COM around the hinge, plus rotation of the door around the COM.

Considering the door initially at rest, the sum of the forces must be greater than zero to move it. $$\mathbf F_a + \mathbf R_h = m\mathbf a_{com}$$, where $$\mathbf F_a$$ is the applied force and $$\mathbf R_h$$ is the reaction at the hinge. The later can be taken as a sum of a component along the door (parallel) and another normal to the door.

$$\mathbf R_h = \mathbf R_{h\perp} + \mathbf R_{h\parallel}$$

Considering the applied force $$\mathbf F_a$$ perpendicular to the door, the torque around the COM is: $$\tau = r_h R_{h\perp} + r_a F_a$$, where $$r_h$$ and $$r_a$$ are the distances from the COM to the hinge and $$\mathbf F_a$$ respectively.

As a result of this torque, the door acquires an angular acceleration around the COM:

$$\tau = I\alpha$$

But at the same time, due to the centripetal force $$R_{h\parallel}$$, the COM rotates around the hinge. $$\frac{R_{h\parallel}}{m} = \omega^2 r_h$$ where $$m$$ is the mass of the door and $$\omega$$ is the instantaneous angular velocity around the hinge.

Of course, as the hinge is a constraint of movement, the combination of rotation of the COM around the hinge, plus the rotation of the door around the COM results in a rotation of the door around the hinge.