Chapter 12, Problem 40AP

A stepladder of negligible weight is constructed as shown in Figure P12.40, 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.

Figure P12.40 Problems 40 and 41.

To determine

The tension in the horizontal bar $DE$.

Answer to Problem 40AP

The tension in the horizontal bar $DE$ is $133\text{N}$.

Explanation of Solution

The length of the each ladder is $4.0\text{m}$, mass of the painter is $70.0\text{kg}$, and the distance of the painter from the point $A$ is $3.0\text{m}$.

The following figure shows the force diagram of the ladder $AC$ and $BC$.

Figure-(I)

Formula to calculate the angle made by the ladder with horizontal is,

    $\cos \theta =\frac{1\text{m}}{l}$

Here, $\theta$ is the angle made by the ladder with horizontal and $l$ is the length of the each ladder.

Substitute $4\text{cm}$ for $l$ in the above equation to find $\theta$.

    $\begin{array}{c}\cos \theta =\frac{1\text{m}}{4\text{cm}}\\ \theta =75.52°\end{array}$

Formula to calculate the torque about the point $A$ is,

    $-mgd\cos \theta +{\overrightarrow{n}}_{\text{B}}\frac{l}{2}=0$

Here, $m$ is the mass of the painter, $g$ is the acceleration due to gravity, $d$ is the distance of the painter from $A$ and ${\overrightarrow{n}}_{\text{B}}$ is the reaction force at point $B$.

Substitute $75.52°$ for $\theta$ in the above equation to find ${\overrightarrow{n}}_{\text{B}}$.

    $\begin{array}{c}-mgd\cos \left(75.52°\right)+{\overrightarrow{n}}_{\text{B}}\frac{l}{2}=0\\ {\overrightarrow{n}}_{\text{B}}=\frac{2mgd\cos \left(75.52°\right)}{l}\end{array}$        (I)

Formula to calculate the net torque about the point $C$ in the right half of the ladder is,

    $-T\left(\frac{l}{2}\right)\sin \theta +n_{\text{B}}\left(\frac{l}{4}\right)=0$

Here, $T$ is the tension in the horizontal bar $DE$.

Substitute $\frac{2mgd\cos \left(75.52°\right)}{l}$ for ${\overrightarrow{n}}_{\text{B}}$ and $75.52°$ for $\theta$ in the above equation to find $T$.

    $\begin{array}{c}-T\left(\frac{l}{2}\right)\sin \left(75.52°\right)+\left(\frac{2mgd\cos \left(75.52°\right)}{l}\right)\left(\frac{l}{4}\right)=0\\ T\left(\frac{l}{2}\right)\sin \left(75.52°\right)=\left(\frac{2mgd\cos \left(75.52°\right)}{l}\right)\left(\frac{l}{4}\right)\\ T=\frac{mgd\cos \left(75.52°\right)}{l\sin \left(75.52°\right)}\end{array}$

Substitute $70.0\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^2$ for $g$, $3.0\text{m}$ for $d$ and $4.0\text{m}$ for $l$ in the above equation to find $T$.

    $\begin{array}{c}T=\frac{\left(70.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(3.0\text{m}\right)\cos \left(75.52°\right)}{\left(4.0\text{m}\right)\sin \left(75.52°\right)}\\ =133.002\text{N}\\ \approx 133\text{N}\end{array}$

Conclusion:

Therefore, the tension in the horizontal bar $DE$ is $133\text{N}$.

To determine

The normal force at $A$ and $B$.

Answer to Problem 40AP

The normal force at $A$ is $429\text{N}$, and the normal force at $B$ is $258\text{N}$.

Explanation of Solution

Formula to calculate the net torque about the point $B$ is,

    $-{\overrightarrow{n}}_{\text{A}}\left(\frac{l}{2}\right)+mg\left(\frac{l}{2}-d\cos \theta \right)=0$

Here, ${\overrightarrow{n}}_{\text{A}}$ is the reaction force at point $A$.

Substitute $70.0\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^2$ for $g$, $3.0\text{m}$ for $d$, $75.52°$ for $\theta$ and $4.0\text{m}$ for $l$ in the above equation to find ${\overrightarrow{n}}_{\text{A}}$.

    $\begin{array}{c}\left[\begin{array}{l}-{\overrightarrow{n}}_{\text{A}}\left(\frac{\left(4.0\text{m}\right)}{2}\right)\\ +\left(70.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(\frac{\left(4.0\text{m}\right)}{2}-\left(3.0\text{m}\right)\cos \left(75.52°\right)\right)\end{array}\right]=0\\ {\overrightarrow{n}}_{\text{A}}\left(\frac{\left(4.0\text{m}\right)}{2}\right)=858.288\text{N-m}\\ {\overrightarrow{n}}_{\text{A}}=429.144\text{N}\\ \approx 429\text{N}\end{array}$

Substitute $70.0\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^2$ for $g$, $3.0\text{m}$ for $d$, and $4.0\text{m}$ for $l$ in the above equation to find ${\overrightarrow{n}}_{\text{B}}$.

    $\begin{array}{c}{\overrightarrow{n}}_{\text{B}}=\frac{2mgd\cos \left(75.52°\right)}{l}\\ =\frac{2\left(70.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(3.0\text{m}\right)\cos \left(75.52°\right)}{\left(4.0\text{m}\right)}\\ =257.55\text{N}\\ \approx 258\text{N}\end{array}$

Conclusion:

Therefore, the normal force at $B$ is $258\text{N}$.

To determine

The components of the force acting on the hinge point $C$ in the right half of the ladder.

Answer to Problem 40AP

The horizontal component of the force at hinge point $C$ is $133\text{N}$ to the right, and the vertical component of the force acting on the hinge point in right half of the ladder is $258\text{N}$ in the downward direction.

Explanation of Solution

Formula to calculate the net horizontal forces acting on the right half of the ladder is,

    $R_{\text{x}}-T=0$

Here, $R_{\text{x}}$ is the horizontal component of the force acting on the hinge point $C$./

Substitute $133\text{N}$ for $T$ in the above equation to find $R_{\text{x}}$.

    $\begin{array}{c}{R}_{\text{x}}-\left(133\text{N}\right)=0\\ R_{\text{x}}=133\text{N}\end{array}$

Formula to calculate the net vertical forces acting on the right half of the ladder is,

    ${\overrightarrow{n}}_B-R_{\text{y}}=0$

Here, $R_{\text{y}}$ is the reaction force on the hinge point $C$.

Substitute  $258\text{N}$ for ${\overrightarrow{n}}_B$ in the above equation to find $R_{\text{y}}$.

    $\begin{array}{c}\left(258\text{N}\right)-R_{\text{y}}=0\\ R_{\text{y}}=258\text{N}\end{array}$

Conclusion:

Therefore, the vertical component of the force acting on the hinge point in right half of the ladder is $258\text{N}$ in the downward direction.