## Essential University Physics: Volume 1 (4th Edition)

$\vec{F_1}=\vec{F_2}=6,400 N$
If the force reaches 10 times their body weight, then the overall downward force is given by: $F = 10*9.8 m/s^2*55 kg$ We know that the force is evenly split between the two muscles. Since the forces are equal to each other, we will find only $\vec{F_1}$. $2\vec{F_1}sin(25^{\circ})= 10*9.8 m/s^2*55 kg$ Finally, we obtain: $\vec{F_1}=\vec{F_2}=6,400 N$