A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0^∘below the horizontal.

step 1 In this problem, we're going to talk about static equilibrium. So what we need to remember is th

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an...

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0$^\circ$ below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it ($\textbf{Fig. E11.20}$). The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension $T$ in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam. Figure E11.20 (CAN'T COPY THE FIGURE)

Key Concepts

Reaction Forces at a Pivot

Reactions at a pivot, especially in a frictionless condition, occur as forces that prevent motion at the point of support. They usually have both horizontal and vertical components and are determined by applying the conditions of static equilibrium to the overall structure. Understanding these reaction forces is key to ensuring the stability of the system.

Force Decomposition

Force decomposition involves breaking a force into its components, typically along mutually perpendicular axes such as horizontal and vertical directions. This is necessary when forces are applied at angles, allowing for easier application of equilibrium equations in each directional component.

Free-Body Diagram

A free-body diagram is a simplified representation of an object where all external forces, moments, and reactions are isolated and drawn. It is a crucial tool for visualizing and setting up equations to apply the equilibrium conditions. By clearly displaying force directions and points of application, it aids in the systematic analysis of structural systems.

Torque and Rotational Equilibrium

Torque is the measure of the tendency of a force to rotate an object about a pivot or axis. For an object to be in rotational equilibrium, the sum of the torques about any pivot point must be zero. This concept is essential when determining unknown forces or tensions in systems that involve lever arms and rotational balance.

Static Equilibrium

In statics, a body is in static equilibrium when the sum of all forces and the sum of all torques (moments) acting on it are zero. This ensures that the body remains at rest without linear or rotational acceleration. The concept is fundamental to analyzing structures and mechanical systems where balance is critical.

Transcript

00:01 In this problem, we're going to talk about static equilibrium.

00:03 So what we need to remember is that the sum of the forces acting on a body for it to be in static equilibrium must be equal to zero, as well as the sum of the torques on the body.

00:19 So both must be equal to zero.

00:21 So that's what we're going to apply.

00:23 So what we have in our problem is a beam that is supporting a street lamp, and the beam itself is supported by a cable represented here in the figure in red that makes a right angle with the beam and also a pivot represented here in black okay and our goal is to find given that the length of the beam is 4 .5 meters the distance between the pivot and the center of mass, the center of gravity, i'm sorry, of the beam is two meters, and the distance between the pivot and the place where the force, the cable force is applied, is three meters, and also we have the information that the tension on the cable is equal to five, i'm sorry, not the tension of the cable, this is the force that the street lamp exerts on the beam downwards is equal to in magnitude it's 5 kiloons and i wrote i wrote it with a minus sign here because i'm setting up a coordinate system where the y -axis points upwards and the x -axis to the right okay so and the weight force is equal to minus 1 .4 kiloons in the j -direction okay, so now we can go on and our goal is to find the magnitude of the tension and also the horizontal and vertical components of the force that the pivot exerts on the beam.

02:19 I'm going to call that force p.

02:23 Okay, so i'm going to start with the torque equation.

02:28 So notice that i'm going to draw a force diagram, a divide diagram on the beam.

02:35 So we have first the graph table.

02:38 Force mg acting at a distance of 2 meters from the pivot then we have the tension force acting at a distance of 3 meters from the beam and we have the force from the from the street lamp that that is acting at a distance of 4 .5 meters from the pivot.

03:16 Okay.

03:18 And also there is a force, the pivot force that's pointing somewhere.

03:25 I'm not sure where yet.

03:29 We're going to find it out.

03:30 I'm just going to write here somewhere.

03:33 It's pointing somewhere, basically.

03:35 It has magnitude p.

03:38 And now i can write the total torque...