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Chapter 8

Q. 8.10

DON’T CLIMB THE LADDER

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.

8.17

Step-by-Step

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.