Does it take more energy to open a door when applying force close to the hinge?

Assuming an ordinary hinged door (without any springs), would it take more energy to open it when applying force in the middle of the door (point b), rather than at the end of the door (point a), where the door knob is?

"Opening the door" should be interpreted as accelerating the door to a certain rotational speed.

My own answer is no, since the change in force would be proportional to the distance required to open the door and therefore the total energy would remain the same. • You need to think of where the energy you are giving to the door is going; friction. Friction of hinges is always the same. Friction of air increases with speed. Slow movement -> 0 air drag. That means that you always need to provide the same energy: the amount wasted on hinge friction.
– user
Apr 15 '15 at 18:45
• Ah, "Force Close," the favorite power of sulky teenage Jedi everywhere. Stay out of my room, Aunt Beru! You're not my real aunt, anyway! Apr 16 '15 at 1:11
• Assuming that the hinge has some friction, could it be that the answer depends on the time $t$ in which we are trying to open the door? If $t$ is very small then clearly we must bang harder at B than at A in order to open the door in time $t$. But if $t$ is very large then it seems like the only energy required is that which is dissipated in the hinges, which is the same when pushing at A or B. (As $t$ goes to infinity it seems like the friction becomes more important, because if there is no friction then the energy required goes to zero; just tap very lightly and wait a very long time.) Apr 16 '15 at 5:52
• (In my comment above I'm ignoring air resistance and also ignoring the energy wasted when muscles contract isometrically.) Apr 16 '15 at 6:42
• Theoretically without springs and without friction it takes (almost) zero energy to open the door. Maybe the question needs to rephrased as the energy required to reach a certain rotational speed $\Omega$. Apr 16 '15 at 13:01

You are right.

To open the door during the same time interval (for pushing at $a$ and $b$), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives $$\frac{F_a}{F_b} = \frac{r_b}{r_a}$$ where $r$ is the distance from the point of contact to the axis.

However the distance that the force should be applied at $a$ and at $b$ are related because they are arcs of the same angle $$\frac{r_b}{r_a}=\frac{l_b}{l_a}$$ This implies that $F_a l_a = F_b l_b$ and equal work needs to be done.

• What about the situation where you forcefully push the door open, rather than pushing it with the same acceleration throughtout the whole movement? Apr 15 '15 at 8:56
• the angular acceleration doesn't have to be constant but can change as long as it changes in the same way in both cases. Apr 15 '15 at 9:00
• I suppose there could be a difference in the maximum speed the door reaches before coming to a stop again. This isn't determined by the physics of the door itself but rather by the person opening the door, and hence that is not easily computed. Apr 15 '15 at 11:24
• What is missing here is the part of the applied torque that goes into the inertial forces. Apr 16 '15 at 13:39
• The force needed to accelerate the center of mass. Apr 17 '15 at 17:42

The physics 101 answer is no: it takes more force, but it is compensated by the smaller displacement so the energy stays the same. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied

But let's dig a bit deeper! When you apply the force, you also have some reaction force on the hinges, which generate some friction, which dissipates energy. So if you stay closer to the hinges, the higher force will in the end require a bit more energy.

We can also consider the device generating the force. If it is your arm, then we have another effect: the energy consumed by muscles only to generate some force is (somehow) proportional to that force. This happens because more muscle fibres have to contract to generate an higher force. Indeed if you try to close ten doors applying the force very close to the hinges, you will be much more tired than closing ten doors from the handle.

This is also the case with an electric motor: higher force (torque) requires higher current which leads to higher ohmic losses, so more energy is dissipated.

It is generally better to keep the forces down when possible.

• Comments are not for extended discussion; this conversation has been moved to chat. Apr 16 '15 at 16:02
• Joined this community just to upvote your answer, specifically for your 3rd paragraph. People tend to forget about biomechanics and come up ridiculous statements, like 'holding a dumbbell above your head requires no energy'. Nov 11 '19 at 15:13

Change of energy is work i.e $W = \Delta E$ and here the work done is $$W = \text{Torque} \cdot \text{angular displacement}$$ which is equal in both the cases. The only change is one needs to apply more force to achieve the same amount of torque at a smaller radius. $$\text{Force at "b"} > \text{Force at "a"}$$ but not the work done or the energy spent.

To get to a final rotational velocity $\Omega$ is takes the same amount of energy regadless of where you push. Why? Well the final kinetic energy is $K=\frac{1}{2} I \Omega^2$ (where $I$ is the mass moment of inertia about the hinge) and this value does not depend on where you push.

This is kind of a boring result.

What does differ is how much you need to push, and how much reaction the hinges provide. If you only push for a small period of time providing with an impulse $J=\int F(t)\,{\rm d}t$ then the hinges would develop a reaction impulse of $R$.

\begin{align} J & = \frac{I \Omega}{a} & R = \frac{I \Omega}{a} - m \Omega \frac{\ell}{2} \end{align}

where $a$ is the distance of where I push from the hinges and $\ell$ is the width of the door. Given that $I=m \frac{\ell^2}{3}$ it can be seen that when $a=\frac{2}{3} \ell$ there is no reaction on the hinges. That is called the center of percussion (sweet spot).

Now the impulse $J=F \Delta t$ can be viewed as an average force $F$ applied for a small time $\Delta t$. If the time of pushing is fixed, then $$F = m \frac{\Omega \ell^2}{3 a \Delta t}$$ which means the further away from the hinge the less the force (duh!).

Now if the distance $\delta$ by which the force is applied is fixed ($\Delta t = \frac{\delta}{v} = \frac{\delta}{a \Omega}$) the force is $$F = m \frac{\Omega^2 \ell^2}{3 \delta}$$ which does not depend on the distance $a$. This can be explained since force over distance is work, which goes into kinetic energy, since the goal is the same kinetic energy, it takes the same amount of work. If the distance is fixed then the force must be fixed also to get to the same work.

Appendix

The two equations of motion I used are

• $J-R = m v_{cm} = m (\Omega \frac{\ell}{2} )$
• $(a-\frac{\ell}{2}) J + \frac{\ell}{2} R = m \frac{\ell^2}{12} \Omega$ Figure 1. Door sketch from the top

• There is no reaction. If the door is free floating in space and you hit it at the CoP it will rotate about the end of the door (where the hinges would have been). May 2 '15 at 13:47
• Given a fixed $J$, yes $\omega$ depends on $a$, but given a target $\omega$ then $J$ varies with $a$. This is what I show, for the two ways of resolving $J$ into a force over time. Using the MMOI $I_{cm}$ about the center the formula is $$J = \dfrac{ \left( I_{cm} + m \left( \frac{\ell}{2} \right)^2 \right) \omega}{a}$$ I am not sure where you got your $2a/I+m a^2$ from, May 2 '15 at 13:59
• Because on a constrained body (hinged) there is no effective mass per se. There is an effective inverse mass (called mobility). If $a=0$ then mobility is 0 (nothing moves). mobility is $$m_{eff}^{-1} = \frac{a^2}{I_{cm}+m \left(\frac{\ell}{2}\right)^2} = \frac{a^2}{I}$$ It turns out that when $a^2=\frac{I}{m}$ (a is the radius of gyration) then $m_{eff}^{-1} = m^{-1}$ May 3 '15 at 14:37
• I posted a question trying to clarify some aspects mentioned here, if you can't answer there, can you explain here a) what is a negative R, b) if you are assuming that J applied equals J acquired by the door (L/a), c) if considering J a real, accountable m*v impulse confirms your conclusions, d) clarify the concept of inverse mass, and if you are considering m = I all along the door
– user78409
Jun 7 '15 at 7:29

...would it take more energy to open it when applying force in the middle of the door rather than at the end of the door

When you push a free body it will translate (yellow arrows) and rotate (white arrows) on the center of mass (C, CM, CoM) at b (CM) it will only translate. If it is hinged it can't translate and is forced to rotate on one tip: pushing at a (the handle) you are following the natural outcome. As you get nearer to the hinge there will be an increasing opposing force (red arrows) contrasting the applied force (black arrows).

The resulting Net force (green arrows) is the outcome, which is different at every point, the max resistance being, of course, near/at the hinge.

so your answer is "Yes, it would take more energy"? – JonathanReez

Because of the definition of mechanical work the energy that produces same ω must be the same, but that definition accounts only for the net force, the last proposition warns you: "Notice that only the component of torque in the direction of the angular velocity vector contributes to the work." other components in other directions, or other opposing forces (which you call increased friction) are not accounted for.

The accepted answer does not take into account the real opposing forces:

it can be seen that when a = 2/3 ℓ, there is no reaction on the hinges. That is called the center of percussion (sweet spot). .. If the door is free floating in space and you hit it at the CoP it will rotate about the end of the door (where the hinges would have been). – ja72

That is not true, that urban legend spread by wikipedia is quickly disproved (I imagine) applying his own formulas, which produce his trochoid the only difference is that at CoP the free 'door' describes a common and from there to the tip a prolate trochoid (where the 'kick' on the wrist of a batter is less 'sweet' because there are more vibrations and, above all, is in the opposite direction). But there are always two opposing forces contrasting the motion and not one (R = J - v).

The opposing forces act before you get $J_{door} = L/a$ and can be deduced considering the conservation laws. Lastly, you could open the door near the hinge with same energy if you could use (not a hand or a kick but) a 40-ton tank on a millimeter, but if you managed that unrealistic task, that would bring down the wall, and probably also the house.

• assuming an ideal door, where would the additional energy be spent? On warming the hinges of the door? Jun 9 '15 at 16:15
• @JonathanReez, you can ask a question about that, but because of the W-E theorem, it's unlikely you get a clear answer, even when it is evident. That is "The elephant in the living room" of this thread. You can find some concrete figures here
– user78409
Jun 11 '15 at 8:48
• this is the correct answer: the accepted answer concerns an ideal situation which can never take place in real world, it's misleading. Sep 2 '18 at 13:36

Torque=(R) x (F) Energy req. to rotate=(T).(Theta), Ta=Tb, ONLY Fa is less than Fb. & T at Hinge=0, it'll not rotate there. The force required will increase from A to hinge point. Energy needed hence is const from A to hinge(except hinge point). [Ta means torque applied at A]

If you open a door by pushing it near the hinge, you apply greater force than when you push it near the outside edge, which requires lesser force since the width of the door acts as a lever and force multiplier. As the friction of the hinge and the weight of the door are equal in both cases, and assuming displacement is the same, net energy transferred to the door is the same in both cases, but only if the speed of the door swinging open is the same in both cases.

Kinetic energy = 0.5 * mass * v^2

You could also solve this problem by using torque. Although energy is a scalar and torque is a vector, they are both expressed in newton meters (joules for energy).

Torque = mass of the door * acceleration * lever arm * sine angle of force applied

The length of the lever arm depends on where you push the door. If you assume the acceleration is proportionately greater the shorter the lever arm, you get the same torque and thus the same energy transferred to the door. Depending on your assumptions, the amount of energy transferred to the door needn't differ from pushing it near its edge.

Assuming an ordinary hinged door ($M = 3Kg$, L = 1m), would it take more energy to open it when applying force in the middle of the door (point b: $r=50cm$), rather than at the end of the door (point a $r=100cm$), My own answer is no, since the change in force would be proportional to the distance required to open the door and therefore the total energy would remain the same.

The issue is not so simple: the equation $E = F*d= F_1*r_1 = F_2*r_2$ is valid only with a lever, where forces are balanced across the fulcrum F. (left sketch) In a door, the center of mass is not at the fulcrum, but at half-length, and the effective mass $m°$of the door varies between $M/3$ (at tip) to $\infty$ (at fulcrum), according to the formula of the rotational mass divided by the square of the distance from the fulcrum/hinge (right diagram): $$m° = \frac{[M]=3*l^2=}{3*d^2}= \frac{1}{d^2}$$

Therefore the nearest to the hinge you are applying the force/impulse, the more energy is required to obtain the same result (the same change of momentum), because the effective rotational mass increases as you get near the fulcrum. In a lever, the mass of the arm is not considered as it is constant, on a hinge, the energy required to displace the door at point $b$ is roughly proportional to the increased mass $E = m*a*d$. I said roughly, because applying a force/torque to a pivoted object is more complex than it may seem, because you have to distinguish between fixed load and follower load. But, grossly simplifying, we may say that it takes a lot more energy at point $b$. Friction and wasted energy are just red herrings

If at first you find this hard to believe, just think that if you apply any force bang to the hinge, no matter how great, the door won't budge. (Unless you bring down the wall, of course). Now, this change of result can't be abrupt, must increase gradually, an it does, by the inverse square law.

What about the situation where you forcefully push the door open ( $J=10kgm/s$), rather than pushing it with the same acceleration throughtout the whole movement? – JonathanReez

! Considering an impulse $J$ (any force must be applied for a time, every force is an impulse) greatly simplifies your calcs. Even though the results are some what different, the principle, is un-changed.

Same for the idea that the change in the door's position can't be abrupt as you move the point of application of the force from the hinge further out. The door won't move if you push at the hinge; it will move if you push elsewhere. That is an abrupt change. . – David Z♦ Practically you are affirming that: if you give K J of energy (the question was about energy so we must always condider that) at point A, the door will not move, then, if you apply same force at 0.0001 m from the hinge (B) or at 1 m distance (C) .the door will move with same angular velocity/momentum/energy.

If that is what you meant, it needs no comment.

I am ready to admit this post is incorrect, if someone shows a (any) concrete example (with real figures) of same force applied at B and C with same result.

• Energy we waste to move the door is consumed by frictions. Hinge friction and air drag. Hinge friction is always the same, and air drag is about 0 for low velocities. Therefore, energy consumed is always the same, and irrelevant of the force we apply.
– user
Apr 15 '15 at 18:55
• @JonathanReez: if we're to take at face value, "just think that if you apply any force bang to the hinge, the door won't budge", as being representative of the issue, then the additional energy is lost in your muscles. If the door doesn't move, then no work is done on it at all. However, it's a feature of animal muscles that they dissipate energy as heat even exerting a static force. Apr 15 '15 at 23:25
• @SteveJessop what if we use a counter-weight to open a door (e.g. a free-falling weight attached to the door using a string)? If we pull at the hinges, the door won't budge, and the weight won't lose it's potential energy. But then if we pull at point B, where does that extra energy go? Apr 15 '15 at 23:28
• I am totally with @JonathanReez. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied. If you compare the impulse, you are not comparing the energy, which is what the question is about. Apr 16 '15 at 7:02
• [citation needed] for the claim that $\Delta E=Fd$ is only valid for a lever with balanced forces - actually, never mind, that's simply wrong. It's practically the definition of work as energy transfer. Same for the idea that the change in the door's position can't be abrupt as you move the point of application of the force from the hinge further out. The door won't move if you push at the hinge; it will move if you push elsewhere. That is an abrupt change. In general, I would want a very good explanation before believing this answer. Apr 16 '15 at 12:46