Summing the moments about a point of rotation equals zero Static Equilibrium Summing the moments about a point of rotation equals zero 5/11/2019 Dr. Sasho MacKenzie - HK 376

Static Equilibrium Conditions The sum of the forces acting on an object equal zero. F = F1 + F2 + F3 …. = 0 This means the object does not translate (no linear acceleration or velocity). The sum of the moments acting on an object equal zero. M = F1r1 + F2r2 + F3r3 … = 0 This means the object does not rotate (no angular acceleration or velocity). 5/11/2019 Dr. Sasho MacKenzie - HK 376

The Direction of Moments The sign of a force is not related to the sign of the moment it produces. A negative force can produce a positive (counter-clockwise) moment and a positive force can produce a negative (clockwise) moment. 5/11/2019 Dr. Sasho MacKenzie - HK 376

A is logical because the point is fixed (Does not move). Fy A 200 N 2 m 1 m Y X In the above figure, 2 forces act perpendicular to a door that is rotating about point A. What must the value of Fy equal in order to maintain static equilibrium? We know that the sum of the moments must equal zero. So we pick a point to sum the moments about. The logical choice is A. A is logical because the point is fixed (Does not move). MA = 0 [Moment 1] + [Moment 2] = 0 [Fy(1 m)] - [(200 N)(2 m)] = 0 Fy = (200 N)(2 m) 1 m Fy = 400 N 5/11/2019 Dr. Sasho MacKenzie - HK 376

A 200 N 400 N 2 m 1 m Y X Fs For the same figure in static equilibrium, calculate the shear force (Fs) acting at the hinges. We know that the sum of the forces must equal zero, so we can sum the forces in the Y direction to determine Fs. Note: there are no forces acting in the X direction (no compression forces). Fy = 0 400 N - 200 N + Fs = 0 Fs = - 400 N + 200 N Fs = - 200 N This negative force means that Fs was drawn incorrectly in the diagram above 5/11/2019 Dr. Sasho MacKenzie - HK 376

15 Fm A 45 N 35 N 13 cm 30 cm 65 cm A shot put is being held in static equilibrium. What force (Fm) must be exerted by the deltoids to maintain this state? A 35 N 45 N Fs Fc Fmy Fmx Y X FBD 5/11/2019 Dr. Sasho MacKenzie - HK 376

First we must resolve the force of the deltoid acting at 15 to the humerus into X and Y components. The X component will not produce a moment, but it will produce a compression force. The Y component will produce a moment and a shear force. 15 Fm Fm 15 Fmx Fmy Fmy = (Fm) sin (15) Fmx = (Fm) cos (15) 5/11/2019 Dr. Sasho MacKenzie - HK 376

Deltoid Force with Shot in Hand MA = [(Fm) sin (15) (13)] - [(35) (30)] - [(45) (65)] = 0 Fm = [(35) (30)] + [(45) (65)] sin (15) (13) = 1183 N Deltoid Force without Shot in Hand MA = [(Fm) sin (15) (13)] - [(35) (30)] = 0 Fm = [(35) (30)] sin (15) (13) = 312.5 N A 35 N 45 N Fs Fc Fmy Fmx Y X 5/11/2019 Dr. Sasho MacKenzie - HK 376

Magnitude of Contact Force at Shoulder Holding Shot The contact force is the resultant of the shear and compression forces acting at the joint. Fy = 0 Fs + (Fm) sin (15) - 35 - 45 = 0 Fs = -1183 sin (15) + 35 + 45 Fs = -226.2 N Shear Force Fx = 0 Fc - (Fm) cos (15) = 0 Fc = 1183 cos (15) Fc = 1142.7 N Compression Force Contact Force =  (-226.2)2 + (1142.7)2 = 1164.9 N A 35 N 45 N Fs Fc Fmy Fmx Y X 5/11/2019 Dr. Sasho MacKenzie - HK 376

Magnitude of Contact Force Not Holding Shot Fy = 0 Fs + (Fm) sin (15) - 35 = 0 Fs = -312.5 sin (15) + 35 Fs = -45.9 N Shear Force Fx = 0 Fc - (Fm) cos (15) = 0 Fc = 312.5 cos (15) Fc = 301.8 N Compression Force Contact Force =  (-45.9)2 + (301.8)2 = 305.3 N A 35 N Fs Fc Fmy Fmx Y X 5/11/2019 Dr. Sasho MacKenzie - HK 376