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Chapter 4: Problem 49

Insect Dynamics. The froghoper (Philaenus spumarius), the champion leaper of the insect world, has a mass of 12.3\(\mathrm{mg}\) and leaves the ground (in the most energetic jumps) at 4.0 \(\mathrm{m} / \mathrm{s}\) from a vertical start. The jump itself lasts a mere 1.0 \(\mathrm{ms}\) before the insect is clear of the ground. Assuming constant acceleration, (a) draw a free-body diagram of this mighty leaper while the jump is taking place; (b) find the force that the ground exerts on the frog hoper during its jump; and (c) express the force in part (b) in terms of the froghopper's weight.

Short Answer

Expert verified
The force exerted by the ground is 0.04932 N, about 409.2 times the froghopper's weight.

Step by step solution

01

Understanding the Problem

First, identify the information given in the problem: The mass of the froghopper is 12.3 mg, which we can convert to kilograms as \( 12.3 \times 10^{-6} \) kg. The initial velocity \( v_0 \) is 0 (since the jump starts from rest), and the final velocity \( v \) after 1 ms (\( 0.001 \) s) is 4.0 m/s.
02

Draw the Free Body Diagram

The free-body diagram should show all forces acting during the jump. Include the force the ground exerts upward on the froghopper and the force of gravity acting downward.
03

Calculate the Acceleration

Use the kinematic equation to find acceleration \( a \): \[ v = v_0 + at \]Substitute \( v = 4.0 \, \text{m/s} \), \( v_0 = 0 \, \text{m/s} \), and \( t = 0.001 \, \text{s} \).Solve for \( a \):\[ a = \frac{v - v_0}{t} = \frac{4.0}{0.001} = 4000 \, \text{m/s}^2 \]
04

Calculate the Net Force

Using Newton's Second Law \( F_{\text{net}} = m \cdot a \), calculate the net force. Given \( m = 12.3 \times 10^{-6} \, \text{kg} \) and \( a = 4000 \, \text{m/s}^2 \),\[ F_{\text{net}} = 12.3 \times 10^{-6} \cdot 4000 = 0.0492 \, \text{N} \].
05

Calculate the Weight of the Froghopper

Calculate the weight \( W \) of the froghopper using \( W = m \cdot g \), where \( g = 9.8 \, \text{m/s}^2 \).\[ W = 12.3 \times 10^{-6} \cdot 9.8 = 0.00012054 \, \text{N} \].
06

Find the Force Exerted by the Ground

The net force \( F_{\text{net}} \) is the difference between the force exerted by the ground \( F_g \) and the weight \( W \). Thus,\[ F_g - W = F_{\text{net}} \].\[ F_g = F_{\text{net}} + W = 0.0492 + 0.00012054 = 0.04932054 \, \text{N} \].
07

Express the Force in Terms of Weight

To express \( F_g \) in terms of weight, we divide by the weight obtained in Step 4.\[ \frac{F_g}{W} = \frac{0.04932054}{0.00012054} \approx 409.2 \].So the force is about 409.2 times the weight of the froghopper.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics in Insect Motion
Kinematics is a branch of mechanics that deals with motion without considering the forces causing it. In the context of insect dynamics, like that of the froghopper, kinematics helps us understand how these incredible jumps occur. We analyze parameters such as velocity, acceleration, and time to describe the motion.

For the froghopper, we begin by noting the initial conditions: its initial velocity before the jump is 0, as it leaps from a stationary position. Upon reaching a final velocity of 4.0 m/s in just 1 ms (0.001 s), these values are critical in understanding how the insect can accelerate so quickly.

The equation for constant acceleration, \(v = v_0 + at\), is foundational here. Using this equation, we solve for acceleration \(a\), revealing the incredible speed-up to 4000 m/s虏. This highlights the precision and adaptability of kinematic principles in nature, showcasing how insects achieve such energetic liftoff.
Newton's Second Law in Action
Newton's Second Law of Motion is essential for understanding how forces act upon objects in motion. The law is typically expressed as \(F_{\text{net}} = m \cdot a\), where \(F_{\text{net}}\) represents the net force applied to an object, \(m\) is the mass, and \(a\) is the acceleration.

For the froghopper, determining the net force during the jump involves calculating its mass (12.3 mg converted to kilograms) and using the acceleration derived from kinematics (4000 m/s虏). Applying Newton鈥檚 Second Law, we find that the net force required for the jump is a precise 0.0492 N.

This insight illustrates how small creatures like the froghopper exert powerful forces relative to their size. Despite its minuscule mass, the resulting force is significant, enabling the insect to achieve its famed leaping prowess.
Creating a Free Body Diagram
A Free Body Diagram (FBD) is a graphical illustration that depicts all the forces acting on an object. It is a valuable tool in physics to visually simplify complex motion scenarios.

In the froghopper鈥檚 case, the FBD shows two main forces during the jump:
  • Upward Force: The force exerted by the ground, propelling the froghopper upwards.
  • Downward Force: The gravitational force acting on the insect.
These opposing forces help analyze the dynamics of the jump and are crucial in calculating the net force. This diagram simplifies understanding by showing how much the ground must counteract gravity to achieve the observed motion, emphasizing the froghopper鈥檚 impressive jump mechanics.
Calculating the Jump Force
Force calculation transforms theoretical physics concepts into quantifiable interactions. Here, it's about quantifying how much force the froghopper generates during its jump.

The net force experienced is 0.0492 N, derived using the acceleration and the froghopper's mass. However, to find the actual ground force, we consider the gravitational pull (or weight) acting on the insect, calculated as \(W = m \cdot g = 0.00012054 \, \text{N}\).

The force exerted by the ground \(F_g\) is the sum of the net force and the weight \( (0.0492 + 0.00012054)\, \text{N} = 0.04932054 \, \text{N}\). This figure is expansive when quantified as over 409 times the insect's weight, highlighting the froghopper's astounding leap relative to its mass. This calculation underscores the immense forces insects harness to perform such unique physical feats.

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Most popular questions from this chapter

A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. The light turns green and you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. Draw clearly labeled free-body diagrams for the truck and for the box. Indicate pairs of forces, if any, that are third-law action- reaction pairs. (The bed of the truck is not friction less.)

A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force \(F=40.0 \mathrm{N}\) that is directed at an angle of \(37.0^{\circ}\) below the horizontal and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton's laws to calculate the normal force that the floor exerts on the chair.

A dockworker applies a constant horizontal force of 80.0 \(\mathrm{N}\) to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 \(\mathrm{m}\) in 5.00 \(\mathrm{s}\) . (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?

A small 8.00 -kg rocket burns fuel that exerts a time-varying upward force on the rocket as the rocket moves upward from the launch pad. This force obeys the equation \(F=A+B t^{2}\) . Measurements show that at \(t=0,\) the force is \(100.0 \mathrm{N},\) and at the end of the first 2.00 s, it is 150.0 \(\mathrm{N}\) . (a) Find the constants \(A\) and \(B\) , including their SI units. (b) Find the net force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after fuel ignition. (c) Suppose you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

An advertisement claims that a particular automobile can "stop on a dime." What net force would actually be necessary to stop a \(850-\) -kg automobile traveling initially at 45.0 \(\mathrm{km} / \mathrm{h}\) in a distance equal to the diameter of a dime, which is 1.8 \(\mathrm{cm} ?\)
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