Question 7.2: A painter on a ladder A 75-kg painter stands on a 6.0-m-long......
A painter on a ladder
A 75-kg painter stands on a 6.0-m-long 20-kg ladder tilted at 53° relative to the ground. He stands with his feet 2.4 m up the ladder. Determine the torque produced by the normal force exerted by the painter on the ladder for two choices of axis of rotation: (a) an axis parallel to the base of the ladder where it touches the ground and (b) an axis parallel to the top ends of the ladder where it touches the wall of the house.
Sketch and translate A sketch of the situation with the known information is shown below. The ladder is our system.
Simplify and diagram Four objects interact with the ladder: Earth, the painter’s feet, the wall, and the ground. Our interest in this problem is only in the torque that the painter’s feet exert on the ladder. Thus, we diagram for the ladder and the downward normal force \vec{N}_{\text {P on } \mathrm{L}} that the painter’s feet exert on the ladder. The diagrams are for the two different axes of rotation.
Represent mathematically We use Eq. (7.1) for the torque caused by the force that the painter’s feet exert on the ladder (for simplicity, we write the magnitude of the force as N with no subscripts): \tau=\pm N l \sin \theta
\tau=\pm F l \sin \theta (7.1)
Solve and evaluate (a) For the first calculation, we choose the axis of rotation at the place where the feet of the ladder touch the ground. The torque produced by the normal force exerted by the painter’s feet on the ladder is
\tau=-N l \sin \theta=-\left(m_{\mathrm{p}} g\right) l \sin \theta
= -(75 kg)[(2.4 m)(9.8 N/kg)]sin 37°
=-1100 \mathrm{~N} \cdot \mathrm{m}
Note that the normal force exerted by the painter’s feet on the ladder has the same magnitude as the force mg that Earth exerts on the painter, but it is not the same force, as it is exerted on a different object and is a normal force and not a gravitational force. That normal force tends to rotate the ladder clockwise about the axis of rotation (a negative torque). The force makes a 37° angle relative to a line from the axis of rotation to the place where the force is exerted.
(b) We now choose the axis of rotation parallel to the wall at the top of the ladder where it touches the wall. The torque produced by the normal force exerted by the painter’s feet on the ladder is
\tau=+F l \sin \theta=+\left(m_{\mathrm{p}} \text{g}\right) l \sin \theta
= +(75 kg)[(3.6 m)(9.8 N/kg)]sin 143°
=+1600 \mathrm{~N} \cdot \mathrm{m}n
When we choose the axis of rotation at the top of the ladder, the downward force exerted by the painter’s feet on the ladder tends to rotate the ladder counterclockwise about the axis at the top (a positive torque). This force makes a 143° angle with respect to a line from the axis of rotation to the place where the force is exerted.
Note that the torque depends on where we place the axis of rotation. You cannot do torque calculations without carefully defining the axis of rotation.
Try it yourself: Write an expression for the torque produced by the upward normal force exerted by the floor on the ladder about the same two axes: (a) at the base of ladder and (b) at the top of the ladder.
Answer: (a) The torque will be 0 at the bottom, because the normal force passes through the axis. (b) About the axis at the top of the ladder, the torque is \tau=-(6.0 \mathrm{~m}) N_{\text {Floor on Ladder }} \sin 37^{\circ} \text {. } Note that the floor tends to push the ladder clockwise about an axis through the top of the ladder. The normal force makes a 37° angle relative to a line from the top axis to the place where the force is exerted.
