# Static Equilibrium Physics 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

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Static Equilibrium Physics 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Static Equilibrium Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Sometimes an object is subject to several forces, but it does not accelerate. This is when the object is in equilibrium. We have done problems like this before, but we neglected the rotational motion. To incorporate this, we simply need to add a torque formula to our typical force formulas. Here’s an example: A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. 1.5m 60°

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. 1.5m 60° We need to draw a diagram of all the forces, then write down force and torque equations: T 3500N 2500N HxHx HyHy T=Tension in wire H x and H y are the components of the force that the hinge exerts on the beam.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. 1.5m 60° We need to draw a diagram of all the forces, then write down force and torque equations: T 3500N 2500N HxHx HyHy 30° This could also be sin(60) We will save this equation and come back to it later. T=Tension in wire H x and H y are the components of the force the hinge exerts on the beam.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. 1.5m 60° We need to draw a diagram of all the forces, then write down force and torque equations: T 3500N 2500N HxHx HyHy This could also be cos(60) We will save this equation and come back to it later. T=Tension in wire H x and H y are the components of the force the hinge exerts on the beam. 30°

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. 1.5m 60° We need to draw a diagram of all the forces, then write down force and torque equations: T 3500N 2500N HxHx HyHy T=Tension in wire H x and H y are the components of the force the hinge exerts on the beam. Before we can fill in the torque equation we need to choose a pivot point. A convenient choice is where the hinge attaches to the beam. This simplifies the torque equation because the 2 unknown hinge forces will not create any torque about that point. Also, remember the sign convention – clockwise torques are negative and counterclockwise positive. Pivot point here 30° +

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. 1.5m 60° We need to draw a diagram of all the forces, then write down force and torque equations: T 3500N 2500N HxHx HyHy T=Tension in wire H x and H y are the components of the force the hinge exerts on the beam. Before we can fill in the torque equation we need to choose a pivot point. A convenient choice is where the hinge attaches to the beam. This simplifies the torque equation because the 2 unknown hinge forces will not create any torque about that point. Also, remember the sign convention – clockwise torques are negative and counterclockwise positive. Pivot point here 30° +

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB A uniform beam 4m long and weighing 2500N carries a 3500N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. 1.5m 60° We need to draw a diagram of all the forces, then write down force and torque equations: T 3500N 2500N HxHx HyHy T=Tension in wire H x and H y are the components of the force the hinge exerts on the beam. Before we can fill in the torque equation we need to choose a pivot point. A convenient choice is where the hinge attaches to the beam. This simplifies the torque equation because the 2 unknown hinge forces will not create any torque about that point. Also, remember the sign convention – clockwise torques are negative and counterclockwise positive. Pivot point here Now we can go back and substitute this value into the other equations to find the hinge forces. 30° +

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