# Force required to open a door of volume lxbxh and weight w

A door of volume l x b x h and weight w is hanging vertically on a horizontal hinge. How can I determine the force/torque to open or close this door?

• largest force is when the door is fully open : parallel to the ground so the system you design has to be able to exceed that. Sep 19, 2017 at 12:13
• I am trying to find the formula for F in terms of l,b,h,weight.... Sep 19, 2017 at 12:18
• @SolarMike I interpreted "vertical hinge" as being like a standard door, in which case the force is position independent (and depends mostly on the friction in the hinges) Sep 19, 2017 at 13:21
• I interpreted "hanging" as in hang man... and the answer posted either read it as I did or followed what i thought... Sep 19, 2017 at 13:24
• Then you should be able to adapt that quite nicely.... Sep 19, 2017 at 15:56

Assuming that:

• the door all one material and the weight is evenly distributed
• the door must be opened all the way to horizontal
• you open it by pushing on the edge furthest from the hinge

then the force required is half the weight. In other words if the mass is 10kg, the weight $w = mg = 10kg*9.8ms^{-2} = 98$N, and 49N will be required to open it. If you want a formula for $F$ then it's probably:

$F = mg/2$

If you can't push on the edge, then more force is needed. If you can't push directly upwards, then more force is needed. If there is any friction in the system, slightly more force is needed. If the door doesn't need to go horizontal, slightly less force is needed. If the door has heavy bits at the outside edge, more force is needed. All in all, it would be wise to design the system with a good amount of extra force.

• Why is force half the weight? I am just dabbling in this topic of physics, because I need to build some mathematical model for my project.... I am from electronics background so I might be asking very basic questions.... Sep 19, 2017 at 16:00
• You balance the moments around the hinge. A moment is the force times the distance. The weight acts through the centre of mass, so is at h/2 from the hinge. The force you apply can be wherever you want but is lowest at the outside edge. The moment from force w at distance h/2 is the same as the moment for force w/2 at distance h. So w/2 is enough to counter the weight because it is applied at twice the distance. Sep 19, 2017 at 16:04
• Actually this calculation gives a lower bound to the force needed to a simple door. The actual force needed is higher how much higher is a question that can not be answered with this data alone. I mean there will be friction if not the door would never be still. Sep 20, 2017 at 13:23
• @joojaa That was kind of the point of the last paragraph in my answer... does it need to be clearer? Sep 20, 2017 at 13:40
• @JackB Yes, it needs to be stated in the beginning. The most important part of a model is what is omitted and what simplified. Also, the model is not very good for a motor that needs to open the door that is acting at the rotation at hinge or with a piston for example. Sep 20, 2017 at 15:47