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## Q. 8.10

GOAL Apply the two conditions of equilibrium.

PROBLEM A uniform ladder $10.0 \mathrm{~m}$ long and weighing $50.0 \mathrm{~N}$ rests against a frictionless vertical wall as in Figure $8.17 \mathrm{a}$. If the ladder is just on the verge of slipping when it makes a $50.0^{\circ}$ angle with the ground, find the coefficient of static friction between the ladder and ground.

STRATEGY Figure $8.17 \mathrm{~b}$ is the force diagram for the ladder. The first condition of equilibrium, $\sum \overrightarrow{\mathbf{F}}_{i}=0$, gives two equations for three unknowns: the magnitudes of the static friction force $f$ and the normal force $n$, both acting on the base of the ladder, and the magnitude of the force of the wall, $P$, acting on the top of the ladder. The second condition of equilibrium, $\Sigma \tau_{i}=0$, gives a third equation (for $P$ ), so all three quantities can be found. The definition of static friction then allows computation of the coefficient of static friction.

## Verified Solution

Apply the first condition of equilibrium to the ladder:

(1) $\Sigma F_{x}=f-P=0 \rightarrow f=P$

(2) $\Sigma F_{y}=n-50.0 \mathrm{~N}=0 \rightarrow n=50.0 \mathrm{~N}$

Apply the second condition of equilibrium, computing torques around the base of the ladder, with $\tau_{\text {grav }}$ standing for the torque due to the ladder’s $50.0-\mathrm{N}$ weight:

$\Sigma \tau_{i}=\tau_{f}+\tau_{n}+\tau_{\text {grav }}+\tau_{P}=0$

The torques due to friction and the normal force are zero about $O$ because their moment arms are zero. (Moment arms can be found from Fig. 8.17c.)

$0+0-(50.0 \mathrm{~N})(5.00 \mathrm{~m}) \sin 40.0^{\circ}+P(10.0 \mathrm{~m}) \sin 50.0^{\circ}=0$

$P=21.0 \mathrm{~N}$

From Equation (1), we now have $f=P=21.0 \mathrm{~N}$. The ladder is on the verge of slipping, so write an expression for the maximum force of static friction and solve for $\mu_{s}$ :

\begin{aligned}21.0 \mathrm{~N} &=f=f_{s, \max }=\mu_{s} n=\mu_{s}(50.0 \mathrm{~N}) \\\mu_{s} &=\frac{21.0 \mathrm{~N}}{50.0 \mathrm{~N}}=0.420\end{aligned}

REMARKS Note that torques were computed around an axis through the bottom of the ladder so that only $\overrightarrow{\mathbf{P}}$ and the force of gravity contributed nonzero torques. This choice of axis reduces the complexity of the torque equation, often resulting in an equation with only one unknown.