# Intuitive explanation for why same force applied farther from a hinge causes larger angular acceleration than if applied closer?

A standard example of a problem involving torque is opening a door - the same force F applied far from the hinge causes a larger angular acceleration than if applied close to the hinge.

I always had trouble getting an intuitive explanation for this - when you're close to the hinge and you're pushing the door with all your strength, it barely moves - so where is all the force going? To the hinge perhaps?

Let's model the door with a rod attached to a hinge. Suppose the rod's length is $L$ with mass $M$, distributed evenly. Its moment of inertia is $I = ML^2/3$. You push the rod with force $F_{applied}$ at distance $d$, giving it an angular acceleration $a$. Using $T=Ia$ we get $F_{applied}*d = Ia$, or $F_{applied}=Ia/d = aML^2/3d$.

Now, if you consider each mass element (particle) of mass $m$ in the rod individually, it experiences a tangential force $F_{tan} = m a_{tan}= m a r$, where $r$ is its distance from the hinge, and so the total tangential force summed over all particles is $F_{tan\ total}=\int_0^L (M/L)\ ar\ dr = aM/2L$.

But according to Newton's 2nd law, $F_{tan\ total} =$ Sum of all external forces in the tangential direction. So $F_{tan\ total} = F_{applied} + F_{some\ other}$.

What is this other force, $F_{some\ other} = aM/2L - aML^2/3d$, which depends on d (your point of application of the force)? Where does it come from?

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Thanks for you answer Mark! Please also see my other question - physics.stackexchange.com/questions/65703/… – Tim May 24 '13 at 3:50

The only error you have is a mistake in your integral. The net force on the door is $\frac{aML}{2}$ in your notation. (Note: it is probably best not to use $a$ for both angular and linear acceleration. $\alpha$ is common for angular acceleration.)
For a fixed applied force, if you eliminate $a$ from your expression, you find that for an applied for $f$, the force from the hinge is $f_{hinge} = f(\frac{3d}{2L} - 1)$. In the limit $d\to 0$, you are pushing right at the hinge and it exactly cancels your push. If you push at $d = \frac{2}{3}L$, there is no force on the hinge. This is the spot where, if the door were floating freely in space, pushing there would move the door forwards, but start it spinning such that the edge of the door stays stationary. If you push at $d=L$, the hinge actually pushes along with you. Since the opposing force from the hinge decreases with increasing $d$, it gets easier to push the door the further out you go.