calculating the force of applied by a vise jaw

garythetiler

Its been many years since i studied mechanics at school and I need to know how to calculate the force applied to a vice jaw by turning the vice handle .i am guessing that it could be worked out by measuring the length oF the threaded bar and also the length from the pivot on the handle but havent got much of a clue after that any help would be appreciated

klawlor419

Can you explain more, where you are trying to measure the force? Do you have schematics of the vice jaw? More details are needed if you want to get a theoretical estimation of this.

Your best bet is to probably using something like this http://www.phidgets.com/products.php?product_id=1131 and determine how turning the vice handle affects the amount of force the vice is exerting experimentally.

Studiot

Hello gary.

The force exerted by the vise is not determined by the vise it is determined by the 'resistance to squashing' of the object between the jaws.

If you place a brick between the jaws you will have to crank the handle very hard to obtain even a small squashing.

On the other hand, if you replace the brick with a sponge car cleaner you can squash the sponge easily.

Objects placed between the jaws push back with a a resistance force that depends upon a property called stiffness and also the amount of squash you give them.

If we consider a fixed amount of squash, say 1mm, then to squash a brick 1mm requires considerably more force than for a sponge. The force equals the stiffness times the squash. Or the stiffness can be called the amount of force needed to squash the object 1mm and is measured in force units (Newtons) per mm.

Now 1mm closure of the vise jaws is determined by the pitch of the leadscrew. This determines how many turns (or part turns) you need to make to close the jaws 1mm.
In turn the length of the handle determines how hard you have to pull on the handle to close the jaws 1mm.

Either way the number of turns to squash a brick is the same as the number of turns to squash a sponge 1mm. The difference is the force you have pull on the handle.
The difference at the other end is the force applied to the object, which depends upon the stiffness of the object as already discussed.

Does this help?

HallsofIvy

Again, "conservation of energy". The total energy applied in turning the handle is the force times the distance the handle has moved through ("work= force times distance"). That same amount of work (energy) is applied to the object and the force applied is that work divided by the distance the jaws have moved.