Question 7.13: Don’t let your ladder slip Any time you have to climb a ladd......
Don’t let your ladder slip
Any time you have to climb a ladder, you want the ladder to remain in static equilibrium. At what angle should a 60-kg painter place his ladder against the wall in order to climb two-thirds of the way up the ladder and have the ladder remain in static equilibrium? The ladder’s mass is 10 kg and its length is 6.0 m. The exterior wall of the house is very smooth, meaning that it exerts a negligible friction force on the ladder. The coefficient of static friction between the floor and the ladder is 0.50.
Sketch and translate We’ve sketched the situation below. If the ladder is tilted at too large an angle from the vertical, everyday experience indicates that it will slide down the wall. The static friction force exerted by the floor on the ladder is what is preventing the ladder from sliding, so one way to look at the situation is to ask, What is the maximum angle relative to the wall that the ladder can have before the static friction force is insufficient to keep the ladder in static equilibrium? We choose the ladder and the painter together as the system of interest.
Simplify and diagram Model the ladder as a rigid body and the painter as a point-like object. A force diagram for the ladder-painter system includes the following forces: the gravitational force exerted by Earth on the ladder \vec{F}_{\text {E on L }} , the gravitational force that Earth exerts on the painter \vec{F}_{\mathrm{E} \text { on }} \mathrm{p} , the normal force of the wall on the top of the ladder \vec{N}_{\mathrm{W} \text { on } \mathrm{L}} \text {, } the normal force of the floor on the bottom of the ladder \vec{N}_{\mathrm{F} \text { on } \mathrm{L}} , and the static friction
force of the floor on the bottom of the ladder \vec{f}_{\text {sF on } L} . We place the axis of rotation where the ladder touches the floor. Rather than determine the center of mass of the system, we have kept the gravitational forces exerted by Earth on the ladder and painter separate.
Represent mathematically Use the force diagram to apply the conditions of equilibrium. Since the forces in this situation point along more than one axis, we use both components of the force condition of equilibrium.
y-component force equation: \Sigma F_y=N_{\mathrm{F} \text { on } \mathrm{L}}+\left(-m_{\text {Ladder }} \text{g}\right)
+\left(-m_{\text {Painter }} g\right)
=0
x-component force equation: \Sigma F_x=\left(-N_{\mathrm{W} \text { on } \mathrm{L}}\right)+\left(f_{\mathrm{s\text{ }F} \text { on } \mathrm{L}}\right)=0
We can insert the expression for the maximum static friction force \left(f_{\mathrm{s} \text { Max F on L }}=\mu_{\mathrm{s}} N_{\mathrm{F} \text { on L }}\right) into the x-component force equation to get
N_{\mathrm{W} \text { on L }}=\mu_{\mathrm{s}} N_{\mathrm{F} \text { on L }}
From the y-component equation, we get
N_{\mathrm{F} \text { on } \mathrm{L}}=m_{\text {Ladder }} \text{g}+m_{\text {Painter }} \text{g}
Combining the last two equations, we get
N_{\mathrm{W} \text { on } \mathrm{L}}=\mu_{\mathrm{s}}\left(m_{\text {Ladder }}+m_{\text {Painter }}\right) \text{g}
The torque equation is
\left[-F_{\mathrm{E} \text { on } \mathrm{L}}\left(\frac{l_{\text {Ladder }}}{2}\right) \sin \theta\right]+\left[-F_{\mathrm{E} \text { on } \mathrm{P}}\left(\frac{2 l_{\text {Ladder }}}{3}\right) \sin \theta\right]
+\left[N_{\text {W on } \mathrm{L}} l_{\text {Ladder }} \sin \left(90^{\circ}-\theta\right)\right]=0
\Rightarrow-m_{\text {Ladder }} \text{g} \frac{l_{\text {Ladder }}}{2} \sin \theta-m_{\text {Painter }} \text{g} \frac{2 l_{\text {Ladder }}}{3} \sin \theta
+\mu_{\mathrm{s}}\left(m_{\text {Ladder }}+m_{\text {Painter }}\right) \text{g} l_{\text {Ladder }} \cos \theta=0
Solve and evaluate We can cancel \text{g} l_{\text {Ladder }} out of each
term of the torque equation and then solve what remains for θ:
\Rightarrow \frac{m_{\text {Ladder }}}{2} \sin \theta+\frac{2 m_{\text {Painter }}}{3} \sin \theta
-\mu_{\mathrm{s}}\left(m_{\text {Ladder }}+m_{\text {Painter }}\right) \cos \theta=0
\Rightarrow\left(\frac{m_{\text {Ladder }}}{2}+\frac{2 m_{\text {Painter }}}{3}\right) \frac{\sin \theta}{\cos \theta}-\mu_{\mathrm{s}}\left(m_{\text {Ladder }}+m_{\text {painter }}\right)=0
\Rightarrow\left(\frac{m_{\text {Ladder }}}{2}+\frac{2 m_{\text {painter }}}{3}\right) \tan \theta=\mu_{\mathrm{s}}\left(m_{\text {Ladder }}+m_{\text {Painter }}\right)
\Rightarrow \tan \theta=\frac{\mu_s\left(m_{\text {Ladder }}+m_{\text {Painter }}\right)}{\frac{m_{\text {Ladder }}}{2}+\frac{2 m_{\text {Painter }}}{3}}
=\frac{6 \mu_{\mathrm{s}}\left(1+m_{\text {Painter }} / m_{\text {Ladder }}\right)}{3+4 m_{\text {Painter }} / m_{\text {Ladder }}}
=\frac{6(0.50)(1+6)}{3+4(6)}=\frac{7}{9}
⇒θ = 38°
This seems reasonable and similar to what we observe in real life when someone uses a ladder. Some ladders come with a warning not to have the angle exceed 15°. As a limiting case, if the coefficient of static friction were zero, then our result says the angle would also have to be zero. This makes sense; without friction between the ground and the ladder, the ladder would need to be perfectly vertical or it would slip.
Try it yourself: If the exterior wall of the house were not smooth, how would the above method need to be changed?
Answer: The wall will exert an upward friction force on the ladder. We would have to take this force into account for both force and torque conditions of equilibrium.