A uniform, 8.0-m 1500-kg beam is hinged to a wall and...

A uniform, $8.0-\mathrm{m}$ $1500-\mathrm{kg}$ beam is hinged to a wall and supported by a thin cable attached 2.0 $\mathrm{m}$ from the free end of the beam, (Fig. P11.50). The beam is supported at an angle of $30.0^{\circ}$ above the horizontal. (a) Draw a free- body diagram of the beam. (b) Find the tension in the cable. (c) How hard does the beam push inward on the wall?

A uniform, $8.0-\mathrm{m}$ $1500-\mathrm{kg}$ beam is hinged to a wall and supported by a thin
cable attached 2.0 $\mathrm{m}$ from the free end of the beam, (Fig. P11.50). The beam is supported
at an angle of $30.0^{\circ}$ above the horizontal. (a) Draw a free- body diagram of the beam. (b) Find the tension in the cable. (c) How hard does the beam push inward on the wall?

Key Concepts

Static Equilibrium

Static equilibrium involves an object being at rest, meaning that both the net force and net torque (moment) acting on the object are zero. This concept is fundamental in mechanics for solving problems where an object is subjected to multiple forces, ensuring that the sum of all forces in any direction and the sum of all moments about any point vanish.

Trigonometric Decomposition of Forces

When forces act at angles, it is necessary to decompose them into perpendicular components (typically horizontal and vertical). This decomposition allows the use of equilibrium conditions by balancing forces in each of the coordinate directions, facilitating the solution of problems involving inclined forces.

Center of Mass

The center of mass of an object is the point where its mass can be considered to be concentrated. In uniform objects, this is at the geometric center. Understanding where the weight of an object effectively acts is essential for accurately calculating the torques in equilibrium problems.

Free-Body Diagrams

A free-body diagram is a simplified representation of an object where all the external forces acting on it are shown. In problems involving static equilibrium, drawing an accurate free-body diagram is crucial as it allows you to visualize forces such as gravity, tension, normal/horizontal forces, and reaction forces, which must be accounted for in the analysis.

Torque or Moment of Force

Torque is the rotational effect produced by a force acting at a distance from a pivot or hinge. It is calculated as the product of the force and the perpendicular distance from the pivot. In equilibrium problems, the balancing of torques is as important as the balancing of linear forces to prevent rotational motion.

Transcript

00:01 Here in this given problem first of all we draw the free body diagram of the uniform beam.

00:16 This is the uniform beam as it is uniform.

00:19 So, its weight should be acting at its center of mass as its mass is 1000.

00:27 So, weight 1000 g acting vertically down.

00:31 This point is marked as c for the center of mass.

00:35 The beam is supposed to be represented as ab and it's making an angle of 30 degree with the horizontal.

00:45 Then there is a rope fixed at a point 2 meter before the other end means this distance that is 2 .00 meter.

01:07 Length of the beam that is given as 8 length means ab that is 8 .0 meter.

01:17 So, ac will be equal to bc means that is half of the length 4 .0 meter.

01:25 Suppose tension in this cable that is t and cable is making an angle 40 degree with the beam.

01:39 Now, we can resolve this 1000 g into two components perpendicular to the rod perpendicular to the beam that is 1000 g cos 30 degree and along the beam this is 1000 g sin 30 degree.

02:05 Then we can resolve this tension t also one of the component perpendicular to the beam if this angle is 40 degree and yes.

02:24 So, the component this component will be t sin 40 degree and along the beam this one that will be t cos 40 degree.

02:37 So, in the first part of the problem free body diagram of the beam that is as shown in the adjoining figure.

03:05 Now, in the second part of the problem we have to find tension in the cable for which we consider rotational equilibrium of the beam about point a.

03:17 Here two more things we should mention here at this end a the force applied by the wall on the beam this is the horizontal component fx and this is the vertical component fy.

03:32 This finishes the free body diagram of the beam...

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