1. **State the problem:** Calculate the net torque acting on a door with width $L=0.6$ m subjected to 6 forces each of magnitude $F=50$ N as shown in the diagram. The hinge is on the left side, and torque direction is defined as + out of the plane and - into the plane. 2. **Recall torque formula:** Torque $\tau = r \times F = rF \sin(\theta)$ where $r$ is the lever arm distance from the hinge, $F$ is the force magnitude, and $\theta$ is the angle between force and lever arm. 3. **Identify forces and lever arms:** - Forces $F_3$ and $F_4$ act at $L/4 = 0.6/4 = 0.15$ m from hinge. - Forces $2F$ act at $L/2 = 0.3$ m from hinge. - Angles for $F_3$ and $F_4$ are $30^\circ$. 4. **Calculate individual torques:** - $\tau_3 = 0.15 \times 50 \times \sin(30^\circ) = 0.15 \times 50 \times 0.5 = 3.75$ Nm - $\tau_4 = 0.15 \times 50 \times \sin(30^\circ) = 3.75$ Nm - $\tau_2 = 0.3 \times 2 \times 50 = 0.3 \times 100 = 30$ Nm - $\tau_5 = 0.3 \times 2 \times 50 = 30$ Nm 5. **Determine torque directions:** - $\tau_3$ and $\tau_4$ produce torque into the plane (negative). - $\tau_2$ and $\tau_5$ produce torque out of the plane (positive). 6. **Calculate net torque:** $$\tau_{net} = (\tau_2 + \tau_5) - (\tau_3 + \tau_4) = (30 + 30) - (3.75 + 3.75) = 60 - 7.5 = 52.5$$ Nm 7. **Final answer:** The net torque is $+52.50$ Nm (positive sign means torque out of the plane).