A ladder with a mass of \(15 \mathrm{~kg}\) rests against a smooth wall ( \(\triangle\) Figure 8.9a). A painter who has a mass of \(78 \mathrm{~kg}\) stands on the ladder as shown in the figure. What is the magnitude of frictional force that must act on the bottom of the ladder to keep it from slipping?

THINKING IT THROUGH. Here there are a variety of forces and torques. However, the ladder will not slip as long as the conditions for static equilibrium are satisfied. Summing both the forces and torques to zero should enable us to solve for the necessary frictional force. Also, as will be seen, choosing a convenient axis of rotation, such that one or more \(\tau\) 's are zero in the summation of the torques, can simplify the torque equation.

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wm 1 y = 5.6 m Nw Ng fs x = 1.0 m x2=1.6 m (a)