Since mechanical advantage is concerned with force, shouldn't the mechanical advantage of a wedge be calculated using the length of the wedge that's parallel to the applied force rather than the length along the sloping side of the wedge?

I've been trying to figure this out, and to make matters more confusing, some sources actually do say that the mechanical advantage is calculated using the length down the middle rather than along the slope. Some sources aren't even consistent with themselves and seem to conflate those two dimensions, as though they've never heard of trigonometry.

I'd be happy with links if anyone can link me to a website that explains this adequately. Thanks!

  • 2
    $\begingroup$ The section found in this book starting from page 202 thru 204, is to be trusted. $\endgroup$ – Novice C Feb 26 '16 at 1:26
  • $\begingroup$ If you think of a wedge as a type of lever, then it is the length of the face which is the lever arm. As you push something up the ramp, the lifting distance is the height of the wedge. $\endgroup$ – Peter Diehr Feb 26 '16 at 2:54

The mechanical advantage is equal to $\dfrac{\text{load “lifted”}}{\text{effort applied}}$

Consider an inclined plane with no frictional forces acting.

enter image description here

The effort needed to push a weight, $mg$, up the slope is $mg \sin \theta$.

So the mechanical advantage of an inclined plane is $\dfrac{mg}{mg \sin \theta} = \dfrac {1}{\sin \theta} = \dfrac{L}{h}$


With wedges there are a number of approximations which have to be made and if one of them is that is the angle of the wedge $\theta$ is small then $\sin \theta \approx \tan \theta$ and this approximation is equivalent to assuming that the hypotenuse is approximately equal to the adjacent side.

For the wedge which is used for splitting wood there is a similar analysis except that the “load” force is at right angles to the surface of the wedge .
Assume there is no friction then the forces which act on the wedge are shown below.

enter image description here

The effort is the force on the wedge $E$ and $G$ the force on the wedge due to the wood which is being split which is equal and opposite to the force exerted by the wedge to split the wood.

$E = 2G \sin \theta \Rightarrow$ mechanical advantage $ = \dfrac G E = \dfrac {1}{2 \;sin \theta}$

Referring to the diagram $\sin \theta = \dfrac {h/2}{L} \Rightarrow $ mechanical advantage $ = \dfrac L h$.

You may wonder about the position of the load force but if you examine a wedge which is used to split wood (right hand diagram) and the head of an axe they both have concave surfaces.
So the position of the load force is a reasonable assumption but it does mean that the mechanical advantage changes a little as the penetration of the wedge increases.

There are other forces in action which have been ignored.
The force on the tip of the wedge due to the wood and also the frictional forces.
For an axe that friction force is reduced by making its surface as smooth as possible.
On the other hand a wedge which is used to split wood is designed not to pop out between being hit by a hammer by having a surface which is ribbed.

All in all the theoretical value of the mechanical advantage of a wedge does give an order of magnitude for its force multiplication but not really much more.

  • $\begingroup$ This is good information, so thank you. But I was hoping for an answer that deals with wedges specifically as opposed to inclined planes, the difference being that wedges are used in motion to separate two things. I am told that mechanical advantage is still the same in both cases, but I'm having trouble understanding why. $\endgroup$ – Kyle Delaney Feb 29 '16 at 0:46
  • $\begingroup$ @KyleDelaney I have added a little about wedge to my answer. $\endgroup$ – Farcher Feb 29 '16 at 13:22

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