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:
[tex]\tau_F=\tau_J\\\\F_Fd_1=F_Jd_2[/tex] (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:
[tex]F_J=\frac{F_Fd_1}{d_2}=\frac{(10.0N)(0.0850m)}{0.027m}=31.48N[/tex]
The force exerted by each jaw is 31.48N