- #1
TheKShaugh
- 22
- 0
Hi,
I am trying to find an equation that would give the mechanical advantage of this system:
I am fairly new to this kind of analysis, but my understanding is that to determine the MA I need to consider that:
[tex]W_{in} = W_{out}[/tex]
[tex]F_{in} d_{in} = F_{out} d_{out}[/tex]
[tex]F_{out} / F_{in} = d_{in} / d_{out} = MA[/tex]
So all I need to do to determine the mechanical advantage is to find the ratio of the distances of the applied and effort forces.
d_in:
The crank and large drum act as a wheel and axle, so the distance the large drum travels per full revolution of the crank shaft is:
[tex]\frac{r_{large \, drum}}{r_{crank}}[/tex]
d_out
The movable pulley is going to rise a distance that equals one half the difference in circumferences between the large and small drum:
[tex]\pi(r_{large \, drum} - r_{small \, drum})[/tex]
Which give a rough mechanical advantage (ignoring the levers created by the crank) of:
[tex] 2 \times \frac{r_{large \, drum}}{\pi r_{crank}(r_{large \, drum} - r_{small \, drum})}[/tex]
I have some concerns though, the main one being that this is different from the only equation I could find online, which is:
[tex]\frac{2 \, r_{crank}}{r_{large \, drum} - r_{small \, drum}}[/tex]
Is my analysis correct or have I missed something? Also, if anyone could tell me why the pulley moves only one half the difference between the two radii and not the full difference between the two I would really appreciate it.
Thanks!
I am trying to find an equation that would give the mechanical advantage of this system:
I am fairly new to this kind of analysis, but my understanding is that to determine the MA I need to consider that:
[tex]W_{in} = W_{out}[/tex]
[tex]F_{in} d_{in} = F_{out} d_{out}[/tex]
[tex]F_{out} / F_{in} = d_{in} / d_{out} = MA[/tex]
So all I need to do to determine the mechanical advantage is to find the ratio of the distances of the applied and effort forces.
d_in:
The crank and large drum act as a wheel and axle, so the distance the large drum travels per full revolution of the crank shaft is:
[tex]\frac{r_{large \, drum}}{r_{crank}}[/tex]
d_out
The movable pulley is going to rise a distance that equals one half the difference in circumferences between the large and small drum:
[tex]\pi(r_{large \, drum} - r_{small \, drum})[/tex]
Which give a rough mechanical advantage (ignoring the levers created by the crank) of:
[tex] 2 \times \frac{r_{large \, drum}}{\pi r_{crank}(r_{large \, drum} - r_{small \, drum})}[/tex]
I have some concerns though, the main one being that this is different from the only equation I could find online, which is:
[tex]\frac{2 \, r_{crank}}{r_{large \, drum} - r_{small \, drum}}[/tex]
Is my analysis correct or have I missed something? Also, if anyone could tell me why the pulley moves only one half the difference between the two radii and not the full difference between the two I would really appreciate it.
Thanks!