Cracking Nuts with Pliers: Estimating Hand Force Needed

[Stratosphere]

Homework Statement

The figure shows scale drawing of a pair of pliers being
used to crack a nut, with an appropriately reduced centimeter grid.
Warning: do not attempt this at home; it is bad manners. If the
force required to crack the nut is 300 N, estimate the force required
of the person's hand.( The Handle is 10cm)

Homework Equations

MA=input force/output force

The Attempt at a Solution

Is this even possible to get the Input force without knowing the MA?

 

[mgb_phys]

You need the mechanical advantage - but this is just the ratio of the handle length to the jaw length (a nutcracker is just a see-saw)

 

[Stovebolt]

You don't need mass or acceleration in this case - you are already given the force. For this problem, you need to look at the moment (or torque) applied vs. output.

 

[mgb_phys]

I think the OP is using MA to mean mechanical advantage (actually strictly speaking in this case - it's velocity ratio)

 

[Stratosphere]

Mechanical Advantage is what i need. So this problem doesn't have enough information?

 

[Stratosphere]

The pliers are not moving, so their angular momentum remains constant
at zero, and the total torque on them must be zero. Not only that, but each half of the pliers
must have zero total torque on it. This tells us that the magnitude of the torque at one end
must be the same as that at the other end. The distance from the axis to the nut is about 2.5

cm, and the distance from the axis to the centers of the palm and ngers are about 8 cm. The
angles are close enough to 90  that we can pretend they're 90 degrees, considering the rough
nature of the other assumptions and measurements. The result is (300 N)(2.5 cm) = (F)(8 cm),
or F = 90 N.
183 I found this at the back of the book.I stil don't get it 8*90 =720 and 300*2.5=750

 

[mgb_phys]

You haven't included the drawing - but presumably you can estimate the distances pivot-nut and pivot-hand, the ratio of these is the mechanical advantage (assuming 100% efficency)