Not sure if this is the right spot but i need some Mechanical Advantage help

kotyluck

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Dec 20, 2012
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The distance from the hinge to the bolt holes is 1/2inches. The distance from the hinge to the end of the handle is 7 inches what is the mechanical advantage of the tool?

so far i did MA=LE over LR = 7 divided by .5inches and got 14inches and so i put 14:1 as the answer. Is this correct? its a first class lever

and going with that "what would the effort at the end of the handle need to be in order to cut the bolt?" shear force needed to cut the bolt = 675lbs. not sure the equation or anything at all for that one.
 
The distance from the hinge to the bolt holes is 1/2inches. The distance from the hinge to the end of the handle is 7 inches what is the mechanical advantage of the tool?

so far i did MA=LE over LR = 7 divided by .5inches and got 14inches and so i put 14:1 as the answer. Is this correct? its a first class lever ............. Correct

and going with that "what would the effort at the end of the handle need to be in order to cut the bolt?" shear force needed to cut the bolt = 675lbs. not sure the equation or anything at all for that one.

By using machine, we need to use less force -

mechanical advantage = (Required force at the bolt) / (applied force at the end of the handle)
 
The distance from the hinge to the bolt holes is 1/2inches. The distance from the hinge to the end of the handle is 7 inches what is the mechanical advantage of the tool?

so far i did MA=LE over LR = 7 divided by .5inches and got 14inches and so i put 14:1 as the answer. Is this correct? its a first class lever

and going with that "what would the effort at the end of the handle need to be in order to cut the bolt?" shear force needed to cut the bolt = 675lbs. not sure the equation or anything at all for that one.

"Mechanical advantage" comes from the fact that there is a "conservation of energy" law- there is NO "conservation of force". If you move the handle through an angle θ\displaystyle \theta (radians) then you have moved the handle through a distance of 7θ\displaystyle 7\theta inches. If the force applied at the handle is Fh\displaystyle F_h then you have done 7Fhθ\displaystyle 7F_h\theta work.

Of course, that will move the bolt holes also through an angle of θ\displaystyle \theta radians and through a distance of (1/2)θ\displaystyle (1/2)\theta inches. If the force exerted at the bolt holes is Fb\displaystyle F_b, then the work done is (1/2)Fbθ\displaystyle (1/2)F_b\theta. By "conservation of energy", those are the same: 7Fhθ=(1/2)Fbθ\displaystyle 7F_h\theta= (1/2)F_b\theta and so FbFh=712=14\displaystyle \frac{F_b}{F_h}= \frac{7}{\frac{1}{2}}= 14.

By the way: 7 inches divided by 1/2 inch is 14, NOT "14 inches". The "inches" 'cancel'.
 
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