### Diagram Description and InstructionsThe image is a diagram illustrating forces acting on a forearm, showing an application of Newton's second law in equilibrium. The forearm is shown with three main forces:1. **Tension Force (\( \vec{F_t} \))**: This force is acting horizontally to the right along the forearm.2. **Force (\( \vec{F_s} \))**: This is directed diagonally downward at an angle (\( \theta = 12^\circ \)) from the horizontal.3. **Gravitational Force (\( \vec{F_g} \))**: This force acts vertically downward at point A, near the hand.**Distances:**- Point O is shown on the elbow, with a horizontal distance of 0.080 m from the tension force application point.- Point A is positioned where the gravitational force is applied, at a horizontal distance of 0.290 m from point O.**Task (A):**Write down the two equations that are derived from Newton’s second law for an object in equilibrium using the given variables:- \( F_t = |\vec{F_t}| \)- \( F_s = |\vec{F_s}| \)### ExplanationIn equilibrium, the sum of forces and the sum of torques acting on the arm should be zero. Therefore, when analyzing the system, these two primary equations need to be established:1. **Sum of Forces Equals Zero**: Ensures the forearm remains stationary or moves at a constant velocity. 2. **Sum of Torques Equals Zero**: Ensures the forearm does not rotate, maintaining a balance of rotational forces.Use these principles to express the equilibrium conditions in mathematical terms, incorporating all given forces and angles.

Solution: From the above-given figure, The Free body diagram can be drawn as below: Now, By applying Newton's second law of motion, we have: In two dimension i.e in x-axis and y-axis for the object in equilibrium is given as: As it is in Equilibrium :∑Fx=0; Then, the equation can be written as:|Fs|cosθ- |Ft|cos120=0As it is in Equilibrium :∑Fy=0; Then, the equation can be written as:|Ft|sin120- |Fs|sinθ -Fg=0Where,In both cases:Fg =Force of attraction due to gravity Thus, The above are the two equations

Answered: F, 12° F, F. 0.080 m 0.290 m- P8.7 from…

F, 12° F, F. 0.080 m 0.290 m- P8.7 from Serway and Vuille 9th Ed. A) Write down the two equations that come from Newton's second Law for an object in equilibrium. You should use the variables F; = |F| and F, = |F| in these equations.