For the stepladder shown in Fig. 12-53, sides A C and C E...

For the stepladder shown in Fig. $12-53,$ sides $A C$ and $C E$ are each 2.44 m long and hinged at $C .$ Bar $B D$ is a tie-rod 0.762 $\mathrm{m}$ long, halfway up. A man weighing 854 $\mathrm{N}$ climbs 1.80 $\mathrm{m}$ along the ladder. Assuming that the floor is frictionless and neglecting the mass of the ladder, find (a) the tension in the tie-rod and the magnitudes of the forces on the ladder from the floor at $(\mathrm{b}) A$ and $(\mathrm{c}) E .$ (Hint: Isolate parts of the ladder in applying the equilibrium conditions.)

For the stepladder shown in Fig. $12-53,$ sides $A C$ and $C E$ are each
2.44 m long and hinged at $C .$ Bar $B D$
is a tie-rod 0.762 $\mathrm{m}$ long, halfway up.
A man weighing 854 $\mathrm{N}$ climbs 1.80 $\mathrm{m}$
along the ladder. Assuming that the floor is frictionless and neglecting the
mass of the ladder, find (a) the tension
in the tie-rod and the magnitudes of
the forces on the ladder from the floor at $(\mathrm{b}) A$ and $(\mathrm{c}) E .$ (Hint: Isolate parts
of the ladder in applying the equilibrium conditions.)

Key Concepts

Static Equilibrium

Static equilibrium is the condition where a structure or body remains at rest because the net force and the net moment acting on it are both zero. This principle is fundamental in solving problems where forces and moments from various loads, supports, and connections must balance perfectly to prevent motion.

Free-Body Diagrams

Free-body diagrams are graphical representations that isolate a body or a section of a structure so that all external forces and moments acting on it are clearly identified. They are indispensable tools in mechanics for setting up the equations of equilibrium by providing a clear visualization of where forces are applied and how they interact.

Moment (Torque) Calculation

Moment or torque calculations are used to determine the turning effect of a force about a pivot point. By considering the perpendicular distance from the point to the line of action of the force, moments help in formulating equilibrium equations about a chosen axis or point, which is critical in solving for unknown forces or tensions in structures.

Frictionless Support Conditions

Frictionless support conditions imply that the contact surfaces provide only normal forces without any frictional resistance. This simplifies the analysis in static problems by eliminating the tangential frictional forces, thereby reducing the complexity of force components that need to be considered at the support points.

Hinged Connections and Tie-Rods

Hinged connections allow components to rotate relative to each other without resisting moments, while tie-rods are structural elements that carry tensile or compressive forces to stabilize or connect parts of a structure. A clear understanding of these connection types is crucial when isolating parts of a structure and applying equilibrium conditions to determine the reaction forces and tensions within the system.

Transcript

00:01 In this problem, on the topic of equilibrium and elasticity, we are shown a step ladder with size ac and ce being 2 .44 meters long and hinged at point c.

00:11 We are then told that a bar or a tie rod, which is 0 .762 meters long, labeled bd, is placed halfway up.

00:21 A man with a weight of 854 neutrons climbs 1 .8 meters along the ladder.

00:26 And if we assume that the floor is frictionless and we neglect the mass of the ladder, we are asked to calculate firstly the tension in the tie rod, as well as the magnitudes of the forces on the ladder from the floor, firstly at a and then at e.

00:43 So the diagrams drawn show the forces on the two sides of the ladder, and fa and fe are the forces of the floor on the two feet.

00:53 T is the tension force of the t -w is the force of the man, which is his weight, fh is the horizontal component of the force exerted by one side of the ladder on the other, and fv is the vertical component of that same force.

01:08 Note that the forces exerted by the floor are normal to the floor since the floor is frictionless.

01:15 Also note that the force of the left side on the right and the force on the right on the left are equal in magnitude and opposite direction.

01:23 Now, since the ladder is in equilibrium, the vertical components of the forces on the left side of the ladder must sum to 0.

01:31 So fv plus f a minus w must equal to 0.

01:41 And the horizontal components similarly must sum to 0 as well for the ladder to be in equilibrium.

01:47 So that's t minus f h and that's must equal to zero.

01:53 Now not only do the forces need to sum to 0, but the talks need to sum to 0 as well.

01:58 If we take the origin to be at the hinge and let l be the length of the ladder, then by equating talks, we get f -a times l cosine theta minus w into l -minus d times cosine theta minus the torque due to the tension of the tyrod t into l over 2 times the sine of theta.

02:31 Must equal to 0.

02:34 Now, here we recognize that the man is a distance d from the bottom of the ladder or l minus d from the top.

02:41 And the tie rod is the midpoint of the, is at the mid point of the side.

02:46 Now, the analogous equations for the right side are that f .e minus fv must equal to zero.

03:09 And the horizontal equations, the horizontal forces, rather, fh minus t, must equal to zero as well.

03:19 And the talk equation is f -e times l cosine theta minus t times l over 2, sine of theta, must equal to zero...

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