Question

If the thumb and finger each squeeze with a force Ft = Ff = 10.0 N, what force do the forceps jaws exert on the plastic rod?

Answer 1

Answer:

FJ = 31.48N

Explanation:

In this case you have that the torque exerted by each, thumb and finger, over the point P, is equal and opposite by the torque exerted by each Jaw.

You equal one of the torque produced by your hand, as for example, the torque of the finger, with the torque produced by one jaw:

$\tau_F=\tau_J\\\\F_Fd_1=F_Jd_2$         (1)

FF: finger force = 10.0N

FJ: jaw force = ?

d1: distance from finger to point P = 8.50cm = 0.085m

d2: distance form the border of the Jaws to the point P = 2.70cm = 0.027m

You solve the equation (1) for FJ and replace the values of the other parameters:

$F_J=\frac{F_Fd_1}{d_2}=\frac{(10.0N)(0.0850m)}{0.027m}=31.48N$

The force exerted by each jaw is 31.48N