Mechanical advantage is defined as Force Output/Force Input

For a symmetrical wedge with the length of the slopes being equal and the width being the distance between the end points, the articles define the Mechanical advantage as: length/width

forces in a wedge

I will assume that you are splitting wood using a symmetrical wedge (lengths of the slopes are equal) and driving the wedge into the wood with a hammer vertically starting with the center of the wooden block. Let us say the width is $w$ and the slope length is $l$. The Input vertical force is resulting in two sideward forces on the wood. Each of these sideward forces is causing the separation in the wood. If, based on the above diagram, I assume that each of these output forces is $F_{in} sin \theta$ where $\theta$ is the angle between the vertical direction of $F_{in}$ and the slope, I have

$$\sin \theta = \left(\frac12 w\right)\cdot\frac{1}{l}, $$

Output force, $$F_{out} = 2 F_{in} \cdot \sin \theta = 2 F_{in} \left(\frac12 w\right)\cdot\frac{1}{l} = F_{in} \cdot \frac{w}{l}$$

or $$\frac{F_{out}}{F_{in}} = \frac{w}{l}$$

This is the inverse of the expected mechanical advantage.

When I equate work done, I get $F_{in} \times height = Load moved \times distance$.

Assuming Load moved is $2*F_{out}$ and distance moved is the perpendicular from center on the slope (perhaps, this statement is wrong. Each point along the vertical was moved by a different amount), I have

$$2*\frac{F_{out}}{F_{in}} = 2\frac{\rm height }{\rm perpendicular\,from\,center\,on\,the\,slope} = 2\sin \theta = 2\left(\frac12 w\right)\cdot\frac{1}{l}= w/l$$

But the mechanical advantage of a wedge is mentioned as $l/w$. Why?

  • $\begingroup$ The mechanical advantage is sometimes defined like a lever ratio, which means the ratio of motion, as in $\eta = \frac{\delta_{out}}{\delta_{in}} $ which is reciprocal to $\frac{F_{out}}{F_{in}}$. $\endgroup$ – ja72 Oct 1 '13 at 2:50
  • $\begingroup$ mechanical advantage, MA, and η, efficiency, are two different things. efficiency = Pout/Pin (Power output/Power Input) which is related to MA as MA = η x (vin/vout) because Power output = Fout x vout and Power input = Fin x vin. $\endgroup$ – pran Oct 2 '13 at 18:13
  • $\begingroup$ yes but I never mentioned efficiency. I can choose to use $\eta$ for MA. Please do not confuse the issue. $\endgroup$ – ja72 Oct 2 '13 at 18:17
  • $\begingroup$ I am not sure about your symbols. I do not know what delta out and in are. Could you point me to a reference? That will help. $\endgroup$ – pran Oct 2 '13 at 23:33
  • $\begingroup$ it is ratio of motion. It can be small displacements, or speeds. $\endgroup$ – ja72 Oct 3 '13 at 12:57

Taking $F_{in}$ as the reference, it is $F_{in}= F_{out} \sin(\theta)$, which makes mechanical advantage proportional to $\frac{1}{\sin(\theta)}$, or $\frac{l}{w}$ for small $θ$.

Taking $F_{out}$ as the reference, as you have chosen, it is $F_{out}= F_{in} \cos(θ)$, which makes mechanical advantage proportional to $\frac{1}{\cos(θ)}$, or $\frac{1}{(1-sin^2(θ))}$ where a simple approximation is not so clear.

  • $\begingroup$ Please use math formatting for better readability. $\endgroup$ – ja72 Oct 1 '13 at 2:47
  • $\begingroup$ From what I know, Mechanical Advantage is defined by Fout/Fin where Fin is the applied input force and Fout is the force/load that is moved. So Fin is pretty clear in this case. I am not sure what you are referring to as 'the reference'. Also not sure how Fout = Fin cos θ for my case. Note that θ is the angle between the vertical and the slope. Finally, cos θ is not (1-sin²θ) but a square root of that but I am not sure why you want to go there. $\endgroup$ – pran Oct 2 '13 at 18:05

The idea behind mechanical advantage is that the work put in equals the work you get out, but the force and distance can be different. So you get this equation:

\begin{equation} W_{in} = W_{out} \end{equation}

Then apply the definition of work: \begin{equation} F_{in}d_{in} = F_{out}d_{out} \end{equation}

Solve for mechanical advantage, which is output divided by input, as you said: \begin{equation} MA = \frac{F_{out}}{F_{in}}\ = \frac{d_{in}}{d_{out}}\ \end{equation}

When you push a wedge, the output distance is how much you push apart, so that's the width of the wedge. As for the input, that's the length of the wedge that you push in.


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