Question 2.6: Holmes’s sky dive Michael Holmes (70 kg) was moving downward......
Holmes’s sky dive
Michael Holmes (70 kg) was moving downward at 36 m/s (80 mi/h) and was stopped by 2.0-m-high shrubbery and the ground. Estimate the average force exerted by the shrubbery and ground on his body while stopping his fall.
Sketch and translate
■ Make a sketch of the process.
■ Choose the system object.
■ Choose a coordinate system.
■ Label the sketch with everything you know about the situation.
We sketch the process, choosing Holmes as the system object H. We want to know the average force that the shrubbery and ground S-G exert on him from when he first touches the shrubbery to the instant when he stops. We choose the y-axis pointing up and the origin at the ground where Holmes comes to rest. We use kinematics to find his acceleration while stopping and Newton’s second law to find the average force that the shrubbery and ground exerted on him while stopping him.
Simplify and diagram
■ Make appropriate simplifying assumptions about the process. For example, can you neglect the size of the system object or neglect frictional forces? Can you assume that forces or acceleration are constant?
■ Then represent the process with a motion diagram and/ or a force diagram(s). Make sure the diagrams are consistent with each other.
We model Holmes as a point-like object and assume that the forces being exerted on him are constant so that they lead to a constant acceleration. A motion diagram for his motion while stopping is shown along with the corresponding force diagram. To draw the force diagram we first identify the objects interacting with Holmes as he slows down: the shrubbery and the ground (combined as one interaction) and Earth. The shrubbery and ground exert an upward normal force \vec{N}_{\mathrm{S}-\mathrm{G} \text { on } \mathrm{H}} on Holmes. Earth exerts a downward gravitational force \vec{F}_{\mathrm{E}} \text { on } \mathrm{H} \text {. } The force diagram is the same for all points of the motion diagram because the acceleration is constant. On the force diagram the arrow for \vec{N}_{\mathrm{S}-\mathrm{G} \text { on } \mathrm{H}} must be longer to match the motion diagram, which shows the velocity change arrow pointing up.
Represent mathematically
■ Convert these qualitative representations into quantitative
mathematical descriptions of the situation using kinematics equations and Newton’s
second law for motion along the axis. After you make the decision about the positive and negative directions, you can determine the signs for the force components in the equations. Add the force components (with either positive or negative signs) to find the sum of the forces.
The y-component of Holmes’s average acceleration is
a_y=\frac{v_y^2-v_{0 y}^2}{2\left(y-y_0\right)}
The y-component of Newton’s second law with the positive y-direction up is
a_y=\frac{\sum F_{\text {on } \mathrm{H} y}}{m_{\mathrm{H}}}
The y-component of the force exerted by the shrubbery-ground on Holmes is N_{\text {S-G on } \mathrm{H} y}=+N_{\text {S-G on } \mathrm{H}} and the y-component of the force exerted by Earth is F_{\mathrm{Eon} \mathrm{H} y}=-F_{\mathrm{E} \text { on } \mathrm{H}}=-m_{\mathrm{H}} g \text {. Therefore, }
a_y=\frac{N_{\mathrm{S}-\mathrm{G} \text { on } \mathrm{H} y}+F_{\mathrm{E} \text { on } \mathrm{H} y}}{m_{\mathrm{H}}}=\frac{\left(+N_{\mathrm{S} \text {-G on } \mathrm{H}}\right)+\left(-F_{\mathrm{E} \text { on } \mathrm{H}}\right)}{m_{\mathrm{H}}}=\frac{+N_{\mathrm{S}-\mathrm{G} \text { on } \mathrm{H}}-m_{\mathrm{H}} g}{m_{\mathrm{H}}}
\Rightarrow N_{\mathrm{S}-\mathrm{G} \text { on } \mathrm{H}}=m_{\mathrm{H}} a_y+m_{\mathrm{H}} g
Solve and evaluate
■ Substitute the known values into the mathematical expressions and solve for the unknowns.
■ Finally, evaluate your work to see if it is reasonable (check units, limiting cases, and whether the answer has a reasonable magnitude). Check whether all representations— mathematical, pictorial, and graphical—are consistent with each other.
Holmes’s average acceleration was
a_y=\frac{0^2-(-36 \mathrm{~m} / \mathrm{s})^2}{2(0-2.0 \mathrm{~m})}=+324 \mathrm{~m} / \mathrm{s}^2
Holmes’s initial velocity is negative, since he is moving in the negative direction. His initial position is +2.0 m at the top of the shrubbery, and his final position is zero at the ground. His velocity in the negative direction is decreasing, which means the velocity change and the acceleration point in the opposite direction (positive). The average magnitude of the force exerted by the shrubbery and ground on Holmes is
N_{\mathrm{S}-\mathrm{G} \text { on } \mathrm{H}}=m_{\mathrm{H}} a_y+m_{\mathrm{H}} g=(70 \mathrm{~kg})\left(324 \mathrm{~m} / \mathrm{s}^2\right)+(70 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})= 22,680 kg . m/s² + 686 N = 23,366 N = 23,000 N
The force has a magnitude greater than the force exerted by Earth—thus the results are consistent with the force diagram and motion diagram. The magnitude is huge and the units are correct. A limiting case for zero acceleration gives us a correct prediction the force exerted on Holmes by the shrubbery and ground equals the force exerted by Earth.
Try it yourself: Use the strategy discussed above to estimate the average force that the ground would have exerted on Holmes if he had stopped in a conservative 0.20 m with no help from the shrubbery.