## Physics (10th Edition)

a) $2593N$ b) $2002N$
The forces that act on the system of the bridge and the hiker are: - The bridge's weight $W_b=3610N$. Its center of gravity is at the midpoint between 2 ends. - The hiker's weight $W_h=985N$ - The holding forces from the concrete at the near end $H_n$ and at the far end $H_f$. We assume these forces point upward to balance the weight. Choose the rotation axis to be at the concrete support at the near end, so that $H_n$ creates no torque. If we take the bridge's length to be $l$, then - For $W_b$, the lever arm is $l/2$, and $\tau_b\lt0$ because the torque created is clockwise. - For $W_h$, the lever arm is $l/5$, and $\tau_h\lt0$ because the torque created is clockwise. - For $H_f$, the lever arm is $l$, and $\tau_f\gt0$ because the torque created is counterclockwise. Because the system is in equilibrium, 1) The net torque is zero: $$\tau_f-\tau_b-\tau_h=0$$ $$H_fl-\frac{W_bl}{2}-\frac{W_hl}{5}=0$$ $$H_f-\frac{W_b}{2}-\frac{W_h}{5}=0$$ $$H_f=\frac{3610N}{2}+\frac{985N}{5}=2002N$$ 2) The net force is zero: $$H_n+H_f-W_b-W_h=0$$ $$H_n=W_b+W_h-H_f=2593N$$