Question 2.10: Froghopper jump The froghopper (Philaenus spumarius), an ins......
Froghopper jump
The froghopper (Philaenus spumarius), an insect about 6 mm in length, is considered
by some scientists to be the best jumper in the animal world. The froghopper can jump 0.7 m vertically, over 100 times higher than its length.
The froghopper’s speed can change from zero to 4 m/s in
about 1 \mathrm{~ms}=1 \times 10^{-3} \mathrm{~s}-\text { an acceleration of } 4000 \mathrm{~m} / \mathrm{s}^2 .
The froghopper achieves this huge acceleration using its leg muscles, which occupy 11% of its 12-mg body mass. What is the average force that the froghopper exerts on the surface during the short time interval while pushing off during its jump?
Sketch and translate We construct an initial-final sketch of the process, choosing the froghopper as the system object. We use a vertical y-axis with the positive direction pointing up and the origin of the coordinate system at the surface. We know the froghopper’s acceleration while pushing off.
Simplify and diagram We can make a motion diagram for the pushing off process and a force diagram. The surface S exerts an upward normal force \vec{N}_{\text {S on } F} on the froghopper, and Earth exerts a downward gravitational force \vec{F}_{\mathrm{E}} \text { on } \mathrm{F}^* Note that because the acceleration is upward, the sum of the y-components of the forces must also be upward; thus, the upward normal force exerted by the surface is greater in magnitude than the downward gravitational force exerted by Earth.
The force diagram shows the normal force exerted by the surface on the froghopper \vec{N}_{\text {S on F }} . We are asked to determine the force that the froghopper exerts on the surface \vec{N}_{\mathrm{F} \text { on } \mathrm{S}} According to Newton’s third law, the magnitude of \vec{N}_{\text {S on } \mathrm{F}} will be the same as the magnitude of \vec{N}_{\text {F on S }} \text {. }
Represent mathematically The process occurs along the vertical direction, so we apply the vertical y-component form of Newton’s second law:
a_y=\frac{\sum F_y}{m_{\mathrm{F}}}=\frac{F_{\mathrm{E} \text { on } \mathrm{F} y}+N_{\mathrm{S} \text { on } \mathrm{F} y}}{m_{\mathrm{F}}}=\frac{-m_{\mathrm{F}} g+N_{\mathrm{S} \text { on } \mathrm{F}}}{m_{\mathrm{F}}}
Rearranging the above, we get an expression for the force of the surface on the froghopper while pushing off:
N_{\text {Son } \mathrm{F}}=m_{\mathrm{F}} a_y+m_{\mathrm{F}} \text{g}
Solve and evaluate We can now substitute the known information in the above, converting the froghopper mass into kg:
12 \mathrm{mg}=(12 \mathrm{mg})\left(\frac{1 \mathrm{~g}}{10^3 \mathrm{mg}}\right)\left(\frac{1 \mathrm{~kg}}{10^3 \mathrm{~g}}\right)=12 \times 10^{-6} \mathrm{~kg}
N_{\mathrm{S} \text { on } \mathrm{F}}=\left(12 \times 10^{-6} \mathrm{~kg}\right)\left(4000 \mathrm{~m} / \mathrm{s}^2\right)+\left(12 \times 10^{-6} \mathrm{~kg}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)
=0.048 \mathrm{~N}+0.00012 \mathrm{~N}=0.048 \mathrm{~N} \approx 0.05 \mathrm{~N}
According to Newton’s third law, the magnitude of the force exerted on the surface by the froghopper is also 0.05 N.
The force that the surface exerts on the froghopper is 400 times greater than the force that Earth exerts on the froghopper. Also, notice that the force magnitudes and both the motion and force diagrams are consistent with each other—a nice check on our work.
Try it yourself: Suppose an 80-kg basketball player could push off a gym floor exerting a force (like the froghopper’s) that is 400 times greater than the force Earth exerts on him and that the push off lasted 0.10 s (unrealistically short). How fast would he be moving when he left contact with the floor?
Answer: About 400 m/s (900 mi/h). He would be moving faster if he took longer to push off while exerting the same force.