Question

A stepladder of negligible weight is constructed as shown in Figure P12.56, with A C=B C= . A painter of mass m stands on the ladder a distance d from the bottom. Assuming the floor is frictionless, find

(a) the tension in the horizontal bar D E connecting the two halves of the ladder,

(b) the normal forces at A and B, and

(c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half. Suggestion: Treat the ladder as a single object, but also treat each half of the ladder separately.

Short answer

$\begin{array}{l}a)T=\frac{mgd}{\sqrt{15L}}\\ b)F_{nB}=\frac{mgd}{2L},F_{nA}=mg\left(\frac{2L-d}{2L}\right)\\ c)R_x=\frac{mgd}{\sqrt{15L}}\left(TowardsEast\right)R_y=\frac{mgd}{2L}\left(Downward\right)\end{array}$

Step-by-step solution

Step 1:. From the geometry of the ladder, we can find θ,

$\begin{array}{l}\cos \theta =\frac{0.25L}{L}\\ \cos \theta =0.25\mathrm{...........}\left(1\right)\end{array}$

Noting that the two sides of the triangle are equal (AC=BC), that's why the two base angles are equal.

 Step2: The external forces exerted on the ladder as one object.

Before start solving the parts a to d, we need to find the external forces exerted on the ladder as one object.

From the figure below, it is obvious that there are only three external forces exerted on the ladder.

And we also know that the system is in static equilibrium which means that

$\sum F_y=0$and$\sum \tau =0$

Hence,

$\begin{array}{l}{F}_{nA}+F_{nB}-mg=0\\ F_{nA}=mg-F_{nB}\end{array}$ (2)

And we will find the net torque around point A and we chose counterclockwise to be the positive torque direction.

$\begin{array}{l}\sum {\tau }_A=0\\ F_{nA}\left(0\right)-(mg\times dcos\theta )+(F_{nB}\times 0.5L)=0\end{array}$

Solve for

$\begin{array}{c}0.5L{F}_{nB}=mgdcos\theta \\ F_{nB}=\frac{mgdcos\theta }{0.5L}\mathrm{.........}\left(3\right)\end{array}$

Step 3: Other normal force by plugging the result of (3) into (2).

Plug from (1)

$F_{nB}=\frac{mgd}{2L}\mathrm{..........}\left(4\right)$

Now we can easily find the other normal force by plugging the result of (3) into (2).

Hence,

$\begin{array}{l}{F}_{nA}=mg-\frac{mgd}{2L}\\ F_{nA}=mg(1-\frac{d}{2L})\\ F_{nA}=mg\left(\frac{2L-d}{2L}\right)\mathrm{.............}\left(5\right)\end{array}$

Note that the two results of (3) and (4) are the answer for part (b).

Step 4:  The tension in the horizontal bar D E connecting the two halves of the ladder,

(a)

To find the tension force of the rope between the two sides of the ladder, we need to draw the force diagram of alone, as the author mentioned in the hint.

We will find the torque around point C, for the right side of the ladder, and we chose counterclockwise to be the positive direction.

Noting that we found the tension force arm (and so do for the normal force at B), as you see in the right side of the figure below, to make finding the torque of this tension force around point C easier.

$\begin{array}{l}\sum {\tau }_C=0\\ R_x\left(0\right)+R_y\left(0\right)-(TLsin\theta /2)+\left(F_{nB}Lcos\theta \right)=0\\ -\frac{1}{2}TLsin\theta +F_{nB}Lcos\theta =0\end{array}$

Solve for T

$\begin{array}{l}T=\frac{2F_{nB}Lcos\theta }{Lsin\theta }\\ T=\frac{2F_{nB}cos\theta }{sin\theta }\mathrm{............}\left(6\right)\end{array}$

Find the normal forces at A and B,

(b)

We know, from (1), that $cos\theta =\frac{1}{4}$

This means that$sin\theta =\frac{y}{4}$ , as you see in the right-triangle below.

And we know, from the Pythagorean theorem, that$4^2=y^2+1^2$

So,

$\begin{array}{l}{y}^2=4^2-1^2\\ y=\sqrt{4^2-1^2}y\\ =\sqrt{15}\end{array}$

Therefore,

$sin\theta =\frac{\sqrt{15}}{4}$

Now plug the value of into (5)

$\begin{array}{l}T=\frac{2F_{nB}\times 0.25}{\frac{\sqrt{15}}{4}}\\ T=\frac{2F_{nB}}{\sqrt{15}}\end{array}$

Plug from (3):

$\begin{array}{l}T=\frac{2mgd}{2L\times \sqrt{15}}\\ T=\frac{mgd}{\sqrt{15}L}\mathrm{...........}\left(7\right)\end{array}$

The components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half

(c)

Since the system is in static equilibrium, so

$\sum F_x=0$

Hence,

$\begin{array}{l}{R}_x-T=0\\ R_x=T\end{array}$

Plug from (6);

$R_x=\frac{mgd}{\sqrt{15L}}$ (Toward east)

And since the system is in static equilibrium, so

$\sum F_y=0$

Hence,

$\begin{array}{c}{F}_{nB}-R_y=0\\ R_y=F_{nB}\end{array}$

Using the equation (3),

$R_y=\frac{mgd}{2L}$ (Downward, see the figure)