Figure shows a tensioning device used to tighten the rear stay of the mast of a sailboat. The block and tackle pulls down a rigid bar with two rollers that squeeze together the two branches of the split rear stay. If the angles are as given in the figure, what is the mechanical advantage?
I am tryign to find the distance from the pivot to F' then divide by the distance to F
$$D_f = 72.5cos \theta$$ $$D_f' = 75cos \theta $$
$$3\frac{D_f}{D_f'} = 2.9 $$
I multiply by three because there is three pulleys in the middle
however I get the wrong answer the answer should be around 34.6 what did I get wrong?
update: I was trying:
$$7\frac{72.5}{15} = 33.8$$
wich is getting close to the answer, but not exactly. where 7 is the number of tensions
So, the key question when figuring out mechanical advantage is figuring out the ratio of how much the input rope moves vs. how much the output moves. Then since the work in is the work out, that gives the ratio of the forces.
The block and tackle just gives a ratio of 4 to the rope - note that it’s not the expected 3 since the rope attaches to the moving pulley. If you kept the initial rope still and pulled the rigid bar, count the number of rolls which would need to stretch.
In the left part of the diagram, the main question is what happens as the rigid bar moves. The location of the bar moves up, so more rope gets pulled into the bottom triangle under the rigid bar as that hypotenuse lengthens. On the other, that means there’s less rope in the top triangle, so that triangle shrinks.
The top string makes an angle of $90-35/2=72.5$ degrees with the horizontal. The bottom string is 15 degrees shallower, so 57.5 degrees. When the height of the rigid bar increases by a bit, the hypotenuse of the bottom triangle increases by $\sin(57.5 ^\circ)$. When the top string’s hypotenuse decreases by $\sin(57.5^\circ)$, the top triangle’s height decreases by $\sin(57.5 ^\circ )/\sin(72.5 ^\circ)$. The overall height of the diagram (and so the location of the output) is the sun of the heights of the two triangles, so since moving the down rigid bar by 1 moves the upper triangle higher by $\sin(57.5 ^\circ )/\sin(72.5 ^\circ)$, it moves the final location by $1-\sin(57.5 ^\circ )/\sin(72.5 ^\circ)$
This means that the total height and so the distance moves by the output is $1-\sin(57.5^\circ)/\sin(72.5 ^\circ )$ times as much as the rigid bar’s movement which due to the block and tackle is $1/4$ times as much as the entire structure’s input movement. This gives a total mechanical advantage of $1/(1/4 \cdot (1-\sin(57.5 ^\circ)/\sin(72.5 ^\circ))) = 34.6$