Chapter 12 - Equilibrium and Elasticity - Problems - Page 349: 38a

To find the force on beam A due to its hinge, we can start by drawing a free-body diagram for beam A, showing all the forces acting on it. Since the beam is in static equilibrium, the sum of the forces in the horizontal direction and the sum of the forces in the vertical direction must both be zero. In the vertical direction, there are two forces acting on beam A: its weight, which points downward and has a magnitude of mAg = 54.0 kg × 9.81 m/s² = 530.3 N (where g is the acceleration due to gravity), and a vertical force from the hinge, which we'll call FvA. Since the beam is not accelerating vertically, we have: ΣFy = mAg + FvA = 0 which gives: FvA = -mAg = -530.3 N The negative sign indicates that the hinge is exerting an upward force on the beam, which is necessary to counteract the weight of the beam and keep it in equilibrium. In the horizontal direction, there are also two forces acting on beam A: a horizontal force from the hinge, which we'll call FhA, and a horizontal force from beam B, which we'll call FAB. Since the beams are loosely bolted together and there is no torque of one on the other, we can assume that FAB is perpendicular to beam A and has no horizontal component. Therefore, we have: ΣFx = FhA = 0 which gives: FhA = 0 This means that the hinge is not exerting any horizontal force on beam A, which makes sense since the beams are not rigidly connected and can move independently in the horizontal direction. To summarize, the force on beam A due to its hinge is: FhA = 0 (in the horizontal direction) FvA = -mAg = -530.3 N (in the vertical direction)

To find the force on beam A due to its hinge, we can start by drawing a free-body diagram for beam A, showing all the forces acting on it. Since the beam is in static equilibrium, the sum of the forces in the horizontal direction and the sum of the forces in the vertical direction must both be zero. In the vertical direction, there are two forces acting on beam A: its weight, which points downward and has a magnitude of mAg = 54.0 kg × 9.81 m/s² = 530.3 N (where g is the acceleration due to gravity), and a vertical force from the hinge, which we'll call FvA. Since the beam is not accelerating vertically, we have: ΣFy = mAg + FvA = 0 which gives: FvA = -mAg = -530.3 N The negative sign indicates that the hinge is exerting an upward force on the beam, which is necessary to counteract the weight of the beam and keep it in equilibrium. In the horizontal direction, there are also two forces acting on beam A: a horizontal force from the hinge, which we'll call FhA, and a horizontal force from beam B, which we'll call FAB. Since the beams are loosely bolted together and there is no torque of one on the other, we can assume that FAB is perpendicular to beam A and has no horizontal component. Therefore, we have: ΣFx = FhA = 0 which gives: FhA = 0 This means that the hinge is not exerting any horizontal force on beam A, which makes sense since the beams are not rigidly connected and can move independently in the horizontal direction. To summarize, the force on beam A due to its hinge is: FhA = 0 (in the horizontal direction) FvA = -mAg = -530.3 N (in the vertical direction)