Question

If a 95 kg gymnast is hanging from a high bar. What force (expressed in Newtons) is being exerted on the high bar? Assuming the high bar is pulling back up on the gymnast with 951 N of force and the coach is exerting two horizontal forces on the gymnast to help stabilize them. The first horizontal force is 25 N pushing on the front of the gymnast and the second horizontal force is 30 N pushing on the back of the gymnast. What is the net resultant force acting on the gymnast and at what angle (from the horizontal) is this force acting? In the space provided draw a free body diagram illustrating the external forces acting on the gymnast’s body. this is for my biomechanics class

Answer 1

Force exerted by the gymnast on the high bar = mg = 95*9.81 = 931.95 N
Now High bar is pulling back the gymnast by a force of 951 N
therefore net force in vertical direction (F_Y)= 951- 931.95 = 19.05 N
Net horizontal force on the gymanst (F_X)= 30 -25 = 5 N
Net forces in the vertical and horizontal direction is shown by the green in the figure.
Now their resultant force
F_R = (F_X² + F_Y²)^1/2 = (5² + 19.05²)^1/2 = 19.695 N
And its direction
= tan^-1(F_Y/F_X) = tan^-1(19.05/5) = 75.293 degree from horizontal direction.