Determine the reaction forces at A and B 50 N 50 N 2n 80 N 160 N 240 N 15 N 30 N 60 N We have studied static equilibrium of forces acting at a common point, and static equilibrium of forces not acting at a common point but involving moments. Now, we will study static equilibrium of entire objects, with the underlying assumption that the objects cannot move nor change shape. These objects are called rigid bodies. We will also study reaction forces at supports. F 5m-5m-5m-5m the diagram above, a horizontal beam is subjected to two downward forces. The weight of the beam is considered negligible, although in some later problems, the weight will be included in the solution. At the left end of the beam, A, is a hinge which allows for rotation The hinge is fixed (attached) and does not allow for any up, down or sideways motion. At point B, the beam simply rests on a triangular support. The beam is not attached to th support. The forces applied at Cand D create clockwise moments about the hinge at A. If equilibrium is to be maintained, there must be a counterclockwise moment. This moment is created by an unward reaction force at B, which we shall call FB. Writing the equation for the sum of the moments about point A: FB(15m)- Fc(5m)-Fo(20m) 0 FB(15m)- Fc(5m)+Fo(20m) For example, if FC# 90 Nand FD :30 N, then FB. (90N +4(30N))/3-|70NİFa] Static equilibrium requires that the sum of the forces is zero, also. The two downward forces add up to 120 N, and the upward force at B is 70 N. Thus, the reaction force at the hinge A must be 50 N upward. This can be determined by taking the moments about B: [Assume FA is upward] Fc(10m)-Fo(5m)-Fa(15m) Fc(10m)-Fo(5m) FA(15m) (2Fc- Fo)/3 FA (2(90 N)-30Ny3 FA 50 N FA Determine the reaction forces at A and B 50 N 50 N 2n 80 N 160 N 240 N 15 N 30 N 60 N We have studied static equilibrium of forces acting at a common point, and static equilibrium of forces not acting at a common point but involving moments. Now, we will study static equilibrium of entire objects, with the underlying assumption that the objects cannot move nor change shape. These objects are called rigid bodies. We will also study reaction forces at supports. F 5m-5m-5m-5m the diagram above, a horizontal beam is subjected to two downward forces. The weight of the beam is considered negligible, although in some later problems, the weight will be included in the solution. At the left end of the beam, A, is a hinge which allows for rotation The hinge is fixed (attached) and does not allow for any up, down or sideways motion. At point B, the beam simply rests on a triangular support. The beam is not attached to th support. The forces applied at Cand D create clockwise moments about the hinge at A. If equilibrium is to be maintained, there must be a counterclockwise moment. This moment is created by an unward reaction force at B, which we shall call FB. Writing the equation for the sum of the moments about point A: FB(15m)- Fc(5m)-Fo(20m) 0 FB(15m)- Fc(5m)+Fo(20m) For example, if FC# 90 Nand FD :30 N, then FB. (90N +4(30N))/3-|70NİFa] Static equilibrium requires that the sum of the forces is zero, also. The two downward forces add up to 120 N, and the upward force at B is 70 N. Thus, the reaction force at the hinge A must be 50 N upward. This can be determined by taking the moments about B: [Assume FA is upward] Fc(10m)-Fo(5m)-Fa(15m) Fc(10m)-Fo(5m) FA(15m) (2Fc- Fo)/3 FA (2(90 N)-30Ny3 FA 50 N FA