## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Let $m_s$ be the mass of the student. Let $m_b$ be the mass of the beam. Let $F_2$ be the force exerted by Support 2. To find the force $F_2$, we can consider the net torque about the axis of rotation located at the position of Support 1. $\sum \tau = 0$ $(3.0~m)~F_2 - r_s~m_s~g-r_b~m_b~g = 0$ $(3.0~m)~F_2 = r_s~m_s~g + r_b~m_b~g$ $F_2 = \frac{r_s~m_s~g + r_b~m_b~g}{3.0~m}$ $F_2 = \frac{(2.0~m)(80~kg)(9.80~m/s^2) + (1.5~m)(100~kg)(9.80~m/s^2)}{3.0~m}$ $F_2 = 1010~N$ Support 2 exerts an upward force of 1010 N on the beam. To find the force $F_1$, we can consider the sum of the vertical forces. $\sum F_y = 0$ $F_1+F_2-m_s~g-m_b~g = 0$ $F_1 = m_s~g+m_b~g-F_2$ $F_1 = (80~kg)(9.80~m/s^2)+(100~kg)(9.80~m/s^2)-1010~N$ $F_1 = 750~N$ Support 1 exerts an upward force of 750 N on the beam.