For the stepladder shown in the Figure, sides$\mathit{A}\mathit{C}$and $\mathit{C}\mathit{E}$ are each $\mathbf{2}\mathbf{.44}\mathit{m}$ long and hinged at . Bar is a tie-rod $\mathbf{0}\mathbf{.762}\mathit{m}$long, halfway up. A man weighing $\mathbf{854}\mathbf{\text{N}}$climbs $\mathbf{1}\mathbf{.80}\mathit{m}$ 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

(b) Aand

(c) E . (Hint: Isolate parts of the ladder in applying the equilibrium conditions.)

Figure:

Length of tie rod $=0.762m$

Weight of man $=854N$

Length of ladder $=1.80m$

Write down the equilibrium condition for both the right and left side bars. We get six equations out of which four are for forces and two for torque. Solve those equations to find the tension. Once we get the tension, use that magnitude of force at floor A and E.

Formula:

$$\begin{gathered} \sum F_x=0 \\ \sum F_y=0 \\ \sum \tau =0 \end{gathered}$$

Consider the free body diagram is shown below:

Left side of the ladder:

Vertical forces must be zero:

$F_v+F_a-W=0$

Horizontal forces must be zero:

$T-F_h=0$

Torque must be zero:

role="math" localid="1663820700156" $F_{a}Lcoa\theta -W(L-d)cos\theta -T\left(\frac{L}{2}\right)sin\theta =0$

Here d is the distance of ladder from the bottom so (L-d from the top)

Now consider the right side:

$F_e-F_v=0$

$F_h-T=0$And$F_{e}Lcoa\theta -T\left(\frac{L}{2}\right)sin\theta =0$

Consider the equations for the conditions is shown below:

$$\begin{gathered} F_v+F_a-W=0 \\ T-F_h=0 \end{gathered}$$

$$\begin{gathered} F_{a}Lcoa\theta -W(L-d)cos\theta -T\left(\frac{L}{2}\right)sin\theta =0 \\ {F}_e-F_v=0 \end{gathered}$$

Also:

$F_{e}Lcoa\theta -T\left(\frac{L}{2}\right)sin\theta =0$

First equation is written as follows:

$F_a=W-F_v$

Consider from the fourth equation as:

$F_v=F_e$

Substitute the equations and solve as:

$T-F_h=0$

$$\begin{gathered} (W-F_v)Lcoa\theta -W(L-d)cos\theta -T\left(\frac{L}{2}\right)sin\theta =0 \\ WLcos\theta -F_{e}Lcos\theta -W(L-d)cos\theta -T\left(\frac{L}{2}\right)sin\theta =0 \end{gathered}$$

$F_{e}Lcos\theta -T\left(\frac{L}{2}\right)sin\theta =0$

Last equation gives:

$F_e=\frac{Tsin\theta }{2cos\theta }=\left(\frac{T}{2}\right)tan\theta$

$T=\frac{Wd}{Ltan\theta }$

The lower side of the triangle has a length of 0.381 m and the hypotenuse has a length of 1.22 m

So the vertical length of triangle is $\sqrt{{1.22}^2-{0.381}^2}$=1.16m

$tan\theta =\frac{1.16}{0.381}=3.04$

$$\begin{gathered} T=\frac{\left(854\right)\left(1.80\right)}{\left(2.44\right)\left(3.04\right)} \\ T=207\text{N} \end{gathered}$$

The equation for the vertical component of force is as follows:

$\begin{array}{c}{F}_v=F_e\\ =\left(\frac{T}{2}\right)\tan \theta \\ =\frac{Wd}{2L}\end{array}$

$\begin{array}{c}{F}_a=W-F_v\\ =W\left(1-\frac{d}{2L}\right)\end{array}$

Substitute the values and solve as:

$\begin{array}{c}{F}_a=854\left(1-\frac{1.80}{2\times 2.44}\right)\\ =539\text{N}\end{array}$

Magnitude of force at A due to floor $539\text{N}$.

Consider the formula:

$F_e=\frac{Wd}{2L}$

Substitute the values and solve as:

$\begin{array}{c}{F}_e=\frac{854\times 1.80}{2\times 2.44}\\ =315\text{N}\end{array}$

Magnitude of force at E due to floor $315\text{N}$.