(II) A uniform rod AB of length 5.0 m and mass M=3.8 kg is hinged at A and held in equilibrium b

step 1 This would be essentially our choosing our coordinates. We can essentially just draw the free bu

(II) A uniform rod AB of length 5.0 m and mass M=3.8 kg is...

(II) A uniform rod AB of length 5.0 $\mathrm{m}$ and mass $M=3.8 \mathrm{kg}$ is hinged at $\mathrm{A}$ and held in equilibrium by a light cord, as shown in Fig. $64 .$ A load $W=22 \mathrm{N}$ hangs from the rod at a distance $x$ so that the tension in the cord is 85 $\mathrm{N}$ (a) Draw a free-body diagram for the rod. (b) Determine the vertical and horizontal forces on the rod exerted by the hinge. (c) Determine $d$ from the appropriate torque equation.

(II) A uniform rod AB of length 5.0 $\mathrm{m}$ and mass $M=3.8 \mathrm{kg}$
is hinged at $\mathrm{A}$ and held in equilibrium by a light cord, as
shown in Fig. $64 .$ A load $W=22 \mathrm{N}$ hangs from
the rod at a distance $x$ so that the tension in the cord is 85 $\mathrm{N}$ (a) Draw a free-body
diagram for the rod. (b) Determine
the vertical and horizontal
forces on the rod exerted by the
hinge. (c) Determine $d$ from the
appropriate torque equation.

Key Concepts

Static Equilibrium

Static equilibrium is a fundamental concept in mechanics stating that when an object is at rest, the net force and net torque acting on it are both zero. This principle allows for the systematic analysis of objects and structures by ensuring that all the forces (and their moments) balance out, which is essential in determining unknown forces and distances.

Free-Body Diagram

A free-body diagram is a simplified drawing that represents an object and all the external forces acting on it. It is a critical tool used to visually organize the different forces, including weight, tension, and reaction forces, which in turn simplifies the application of equilibrium equations in solving for unknown quantities.

Torque (Moment) Calculation

Torque, or the moment of force, refers to the rotational effect produced by a force acting at a distance from a pivot point. In equilibrium problems, torque calculations are used to ensure that the sum of the moments about any pivot point is zero, which assists in determining unknown distances or forces acting on the object.

Force Decomposition

Force decomposition involves breaking a force into its horizontal and vertical components. This technique is essential in solving equilibrium problems because it allows for the independent application of the equilibrium conditions in each perpendicular direction, facilitating the calculation of unknown reaction forces or other applied forces.

Transcript

00:01 This would be essentially our choosing our coordinates.

00:06 We can essentially just draw the free buddy diagram and we're going to choose upwards to be positive y to the right to be positive x.

00:14 And then for part b they want us to find the horizontal and vertical components of the hinge force.

00:20 So we can apply the sum of forces in the x direction.

00:23 We know this to be equaling zero and this is going to be equal to the force of the rope.

00:28 Multiplied by sine of phi minus the force of the we can say the hinge horizontal and so the force of the hinge horizontal is going to be equaling the force of the rope times sine of phi and we can then solve the force of the hinge horizontal will be equalling 85 newtons multiplied by sign of 37 degrees.

01:10 This is equaling approximately 51 newtons.

01:14 This would be our answer, our first answer for part b, the horizontal component of the hinge force.

01:20 And then applying some of forces in the y direction, this is also equalling zero.

01:26 The system has static equilibrium as well as translational equilibrium in both the x and y directions.

01:33 And so the sum of forces in the y direction is going to equal force of the rope times cosine of phi plus the vertical force of the hinge so hinge comma vert minus mg minus the weight w this is again equaling zero and so we can then say that the vertical component of the hinge force is equaling m g plus the weight minus force of the rope cosine of phi and so we can solve and say that the vertical component of the hinge force is equaling 3 .8 kilograms multiplied by 9 .8 meters per second squared plus 22 newtons minus 85 newtons times cosine of 37 degrees and our magnitude of the vertical component of the hinge force is going to be negative 8 .6 newtons.

02:56 This would be our answer...

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