A diving board 3.00 m long is supported at a point 1.00 m from the end, and a diver weighing 500 N stands at the free end (Fig. E11.11). The diving board is of uniform cross section and weighs 280 N. Find (a) the force at the support point and (b) the force at the left-hand end.
Solution 34E The system of the person and diving board is at rest so the two conditions of equilibrium apply;F = 0 the vector sum of all external forces acting on the body is zero; =0 the sum of the torques due to all external forces acting on the body, with respect to any specified point, must be zero. Step 1 of 2: The free body diagram for the diving board is as shown in the fig below. Take the origin of the co ordinates at the left hand end of the board. F 1 is the force applied at the support point and F 2 is the force at the end A that is held down. Now, = 0which gives A +F (1.00) (500 × 3.00) (280 × 1.50)= 0 1 (500×3.00) +(280×1.50) F 1 1.00 = 920 N