Question

A 65.0-kg painter is on a uniform 25-kg scaffold supported from above by ropes (Fig. 9–84). There is a 4.0-kg pail of paint to one side, as shown. Can the painter walk safely to both ends of the scaffold? If not, which end(s) is dangerous, and how close to the end can he approach safely?

Short answer

Both ends are dangerous.

Step-by-step solution

Understanding of torque

The term "torque" may be defined as the turn or twist to generate the rotation of a body. It is computed by multiplying the magnitude of the force by the lever arm length.

Given information

Given data:

The mass of the painter,\(M = 65.0\;{\rm{kg}}\).

The mass of the scaffold, \(m = 25\;{\rm{kg}}\).

The mass of the pail of paint, \({m_{\rm{P}}} = 4.0\;{\rm{kg}}\).

Evaluation of the distance to the right that the painter can walk before the tension in the left rope becomes zero

The safety of the painter is dependent on the tension in the left rope. The free-body diagram of the system is as follows:

Take the torques about the right rope, with counterclockwise torques as positive.

\(\begin{array}{c}\sum \tau = 0\\{m_{\rm{P}}}g\left( {3\;{\rm{m}}} \right) + mg\left( {2\;{\rm{m}}} \right) - Mgx = 0\\{m_{\rm{P}}}g\left( {3\;{\rm{m}}} \right) + mg\left( {2\;{\rm{m}}} \right) = Mgx\\x = \frac{{{m_{\rm{P}}}\left( {3\;{\rm{m}}} \right) + m\left( {2\;{\rm{m}}} \right)}}{M}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}x = \frac{{\left( {4\;{\rm{kg}}} \right)\left( {3\;{\rm{m}}} \right) + \left( {25\;{\rm{kg}}} \right)\left( {2\;{\rm{m}}} \right)}}{{\left( {65\;{\rm{kg}}} \right)}}\\x = 0.95\;{\rm{m}}\end{array}\)

This distance is less than the distance to the end of the plank; therefore, walking to the right end is not safe.

Evaluation of the distance to the left that the painter can walk before the tension in the right rope becomes zero

The free-body diagram for the left side is shown below:

Take the torques about the left rope, with counterclockwise torques as positive.

\(\begin{array}{c}\sum \tau = 0\\{m_{\rm{P}}}g\left( {1\;{\rm{m}}} \right) + mg\left( {2\;{\rm{m}}} \right) - Mgx = 0\\{m_{\rm{P}}}g\left( {1\;{\rm{m}}} \right) + mg\left( {2\;{\rm{m}}} \right) = Mgx\\x = \frac{{{m_{\rm{P}}}\left( {1\;{\rm{m}}} \right) + m\left( {2\;{\rm{m}}} \right)}}{M}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}x = \frac{{\left( {4\;{\rm{kg}}} \right)\left( {1\;{\rm{m}}} \right) + \left( {25\;{\rm{kg}}} \right)\left( {2\;{\rm{m}}} \right)}}{{\left( {65\;{\rm{kg}}} \right)}}\\x = 0.8307\;{\rm{m}} \approx {\rm{0}}{\rm{.8}}\;{\rm{m}}\end{array}\)

This distance is less than the distance to the end of the plank; therefore, walking to the left end is not safe.

Thus, both ends are dangerous.