Chapter 10, Problem 73P

A stepladder of negligible weight is constructed as shown in Figure P10.73, with AC = BC = ℓ = 4.00 m. A painter of mass m = 70.0 kg stands on the ladder d = 3.00 m from the bottom. Assuming the floor is frictionless, find (a) the tension in the horizontal bar DE 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.

To determine

The tension experienced on the horizontal bar $DE$.

Answer to Problem 73P

The tension experienced on the horizontal bar $DE$ is $\underset{\_}{133\text{N}}$.

Explanation of Solution

From the geometry of the ladder the angle $\theta$ is the angle inclined by the bottom of the ladder with the frictionless ground will be.

  $\begin{array}{c}\cos \theta =\frac{1}{4}\\ \theta ={\cos }^{-1}\left(\frac{1}{4}\right)\\ =75.5°\end{array}$

First of all consider the net torque at the point A which is at the bottom left side of the ladder. The torque experienced at this point is due to the weight of the painter, and the normal force at point B, $n_B$. The component of weight of the painter at A is $mgd\cos \theta$. And point A and B are separated by a distance $\frac{\mathcal{l}}{2}$.

Write the expression for the torque at point A.

  ${\displaystyle \sum {\tau }_{\text{A}}}=-mgd\cos \theta +n_B\frac{\mathcal{l}}{2}$        (I)

Here, $m$ is the mass of the painter, $g$ is the acceleration due gravity, $d$ is the distance from painter to the bottom of the ladder.

The net torque at point A is zero, hence equate equation (I) to zero and obtain an expression for $n_B$ by substituting $75.5°$ for $\theta$.

  $\begin{array}{c}-mgd\cos 75.5°+n_B\frac{\mathcal{l}}{2}=0\\ n_B\frac{\mathcal{l}}{2}=\frac{\mathcal{l}}{2}mgd\cos 75.5°\\ n_B=\frac{2}{\mathcal{l}}mgd\left(\frac{1}{4}\right)\\ =\frac{mgd}{2\mathcal{l}}\end{array}$        (II)

Now consider the net torque acting at the point B, which is at the bottom right of the ladder. The torque at point B same as A, is due to the weight of the painter and normal force at point A, $n_A$.

Write the expression for the net torque at point B.

  ${\displaystyle \sum {\tau }_B}=-n_A\frac{\mathcal{l}}{2}+mg\left(\frac{\mathcal{l}}{2}-d\cos 75.5°\right)$        (III)

The net torque acting at point B will be zero, equate equation (III) to zero, and obtain an expression for $n_A$.

  $\begin{array}{c}-n_A\frac{\mathcal{l}}{2}+mg\left(\frac{\mathcal{l}}{2}-d\cos 75.5°\right)=0\\ n_A\frac{\mathcal{l}}{2}=mg\left(\frac{\mathcal{l}}{2}-d\cos 75.5°\right)\\ n_A=\frac{2}{\mathcal{l}}mg\left(\frac{\mathcal{l}}{2}-d\cos 75.5°\right)\\ =mg\left(1-\frac{d}{2\mathcal{l}}\right)\end{array}$        (IV)

Consider the torque at the point Cat the top of the right half of the ladder. At this point torque is due to the tension, $T$, and the normal force, $n_B$.

Write the expression for the torque at point C.

  ${\displaystyle \sum {\tau }_C}=-T\frac{\mathcal{l}}{2}\sin 75.5°+n_B\frac{\mathcal{l}}{4}$        (V)

The net torque at point C is zero. Equate equation (III) to zero, and obtain an expression for $T$.

  $\begin{array}{c}-T\frac{\mathcal{l}}{2}\sin 75.5°+n_B\frac{\mathcal{l}}{4}=0\\ T\frac{\mathcal{l}}{2}\sin 75.5°=n_B\frac{\mathcal{l}}{4}\\ T=\frac{\mathcal{l}{n}_B}{4}\times \frac{2}{\mathcal{l}\sin 75.5°}\end{array}$        (VI)

Substitute, equation (II) in (VI).

  $\begin{array}{c}T=\frac{\mathcal{l}}{4}\times \frac{2}{\mathcal{l}\sin 75.5°}\frac{mgd}{2\mathcal{l}}\\ =\frac{mgd}{4\mathcal{l}\sin 75.5°}\end{array}$        (VII)

The tension $T$ will provide a clockwise torque about point C in order to balance the counterclockwise torque from $n_B$. Let $R_x$ be the reaction force acting on the right half of the ladder towards right, and $R_y$ be the reaction force acting on the right half towards downwards.

Write the expression for the force acting along right.

  ${\displaystyle \sum F_x=R_x-T}$        (VIII)

Equate equation (VIII) to zero.

  $\begin{array}{c}{R}_x-T=0\\ R_x=T\end{array}$        (IX)

Write the expression for the force acting along $y$ direction.

  ${\displaystyle \sum F_y=R_y+n_B}$        (X)

Equate equation (VIII) to zero.

  $\begin{array}{c}{R}_y+n_B=0\\ R_y=-n_B\end{array}$        (XI)

From equation (XI) the reaction force $R_y$ is acting downwards.

Conclusion:

Substitute, $70.0\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$, $3.00\text{m}$ for $d$, and $4.00\text{m}$ for $\mathcal{l}$ in equation (VII) to find the tension.

  $\begin{array}{c}T=\frac{\left(70.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(3.00\text{m}\right)}{4\left(4.00\text{m}\right)\sin 75.5°}\\ =133\text{N}\end{array}$

Therefore, the tension experienced on the horizontal bar $DE$ is $\underset{\_}{133\text{N}}$.

To determine

The normal force acting at A, and B.

Answer to Problem 73P

The normal force acting at A is $\underset{\_}{n_A=429\text{N}}$, and at B $\underset{\_}{n_B=257\text{N}}$.

Explanation of Solution

Use equation (II) and (IV) to obtain the answer.

Conclusion:

Substitute, $70.0\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$, $3.00\text{m}$ for $d$, and $4.00\text{m}$ for $\mathcal{l}$ in equation (II) to find $n_A$.

  $\begin{array}{c}{n}_A=\left(70.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(1-\frac{3.00\text{m}}{2\times 4.00\text{m}}\right)\\ =429\text{N}\end{array}$

Substitute, $70.0\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$, $3.00\text{m}$ for $d$, and $4.00\text{m}$ for $\mathcal{l}$ in equation (II) to find $n_B$.

  $\begin{array}{c}{n}_B=\frac{\left(70.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(3.00\text{m}\right)}{2\times 4.00\text{m}}\\ =257\text{N}\end{array}$

Therefore, the normal force acting at A is $\underset{\_}{n_A=429\text{N}}$, and at B $\underset{\_}{n_B=257\text{N}}$.

To determine

The components of reaction forces at point C.

Answer to Problem 73P

The components of reaction forces at point C are, $\underset{\_}{R_x=133\text{N to the right}}$, and $\underset{\_}{R_y=257\text{N, downwards}}$.

Explanation of Solution

Consider equation (IX) and (XI), the $x$ component is equal to the tension, and $y$ component is equal to the normal force.

The tension experienced on the horizontal bar $DE$ is. Hence the reaction force $R_x$ is.

  $R_x=133\text{N}$

The normal force acting at B is $257\text{N}$. Hence the reaction force is.

  $R_y=257\text{N}$

Conclusion:

Therefore, the components of reaction forces at point C are, $\underset{\_}{R_x=133\text{N to the right}}$, and $\underset{\_}{R_y=257\text{N, downwards}}$.