Question is included in image provided **Problem: Torque, Angular Kinematics, Newton’s 2nd Law**- A 2"×4" board of length 2.0 m and mass 5.0 kg is supported by two saw horses which are 1.0 m high, one at each end of the board. (a) If the saw horse on one end is kicked out from under the board, what is the magnitude of the initial acceleration of that end of the board? (b) What force must be exerted on the board by each saw horse for the board to remain in static equilibrium?*Approximate the board as a thin rod for its moment of inertia.***Diagram Explanation:**The diagram illustrates the setup:- A horizontal board is placed on top of two vertical supports (saw horses), depicted as blue lines.- The board is shown as a horizontal line resting on the two supports, each 1.0 m high.- The diagram visually represents the problem scenario to aid in understanding the concepts of torque, equilibrium, and angular kinematics.

Given Information: The mass of the rod (m) = 5 kg The length of the rod (l) = 2 m The height of the saw horse (h) = 1.0 m The cross section of the beam = 2''×4'' If the saw horse is removed, the free body diagram of the board is: The torque on the board is given as: T=mg l2=5(9.81)22=49.05 N The angular acceleration of the board is given as: T=IαThe board rotates with the axis at top of the saw horse.I = ml23=5(2)23=6.66 kgm2The angular acceleration is given as:α=TI=49.056.66=7.36 rad/sec2 The tangential acceleration of the end of the beam is: at=rα=lαat=2(7.36)=14.72 m/s2 Free body diagram for the static equilibrium is: The equilibrium of the equation: The moment about point A must be “Zero”, we can write it as: Now, from the Fy we can write: