back to book: LEANING LADDER

Saturday, March 21, 2015

LEANING LADDER

Worked example 10.3: Leaning ladder

Question: A uniform ladder of mass $m = 40 {\rm kg}$ and length $l=10 {\rm m}$ is leaned against a smooth vertical wall. A person of mass $M=80 {\rm kg}$ stands on the ladder a distance $x=7 {\rm m}$ from the bottom, as measured along the ladder. The foot of the ladder is $d=1.2 {\rm m}$ from the bottom of the wall. What is the force exerted by the wall on the ladder? What is the normal force exerted by the floor on the ladder?

$\begin{figure*} \epsfysize =2.5in \centerline{\epsffile{ladder1.eps}} \end{figure*}$

Answer: The angle $\theta$ subtended by the ladder with the ground satisfies

$\begin{displaymath} \theta = \cos^{-1}(d/l)= \cos^{-1}(1.2/10) = 83.11^\circ. \end{displaymath}$

Let

be the normal reaction at the wall, let

be the normal reaction at the ground, and let

be the frictional force exerted by the ground on the ladder, as shown in the diagram. Consider the torque acting on the ladder about the point where it meets the ground. Only three forces contribute to this torque: the weight,

, of the ladder, which acts half-way along the ladder; the weight,

, of the person, which acts a distance

along the ladder; and the reaction,

, at the wall, which acts at the top of the ladder. The lever arms associated with these three forces are $(l/2) \cos\theta$ , $x \cos\theta$ , and $l \sin\theta$ , respectively. Note that the reaction force acts to twist the ladder in the opposite sense to the two weights. Hence, setting the net torque to zero, we obtain

$\begin{displaymath} m g \frac{l}{2} \cos\theta + M g x \cos\theta - S l \sin\theta = 0, \end{displaymath}$

which yields

$\begin{displaymath} S = \frac{(m g/2 + M g x/l)}{\tan\theta} = \frac{(0.5\tim... ...80\times 9.81\times 7/10)}{\tan 83.11^\circ} = 90.09 {\rm N}. \end{displaymath}$