Question

Try to show all of the following steps for the problem below:

Draw a free-body diagram for at least one object (or system) with vectors showing all forces on that object.

Draw a coordinate system for each free-body diagram and label the axes clearly.

Write Newton’s Second Law in component form for each object (that is: write ∑Fx = max , ∑Fy = may , etc)

Write down any constraints for the system based on what you know about its motion. For instance, you may know that the velocity is zero or that the velocity is constant or that two objects move together or that the motion must lie along a certain axis.

Solve algebraically for the unknowns.

Coal miners often find mice in deep mines but rarely find rats; let’s see if we can figure out why. A mouse is roughly 5 cm long by 2 cm wide and has a mass of 30 g; a rat is roughly 20 cm long by 5 cm wide and has a mass of 500 g. Assume that both have a drag coefficient CD ≈ 0.3.

a) Estimate the terminal falling speeds reached by a mouse and a rat, respectively.

b) Assume that mine shafts are deep enough that both the mouse and the rat reach terminal velocity before hitting the bottom. Estimate the magnitude of the maximum force required to stop both the rat and the mouse when they hit the bottom. (Hint: Over what distance will the center of mass travel between the beginning of the collision and when the animal is at rest? What is the acceleration required to bring the animal to rest over this distance? The center of mass can’t travel more than about half the height of an animal, why?)

c) Bones will break if they are subjected to a compressional force per unit area of more than about 1.5 × 108 N/m2. A mouse may have leg bones about 1.5 mm in diameter; for a rat, they might be about twice as thick. Using your estimates from part b), determine if either the mouse or the rat (or both) will suffer broken legs. Is your answer consistent with the observations of the coal miners? Explain.

Answer 1

we know that terminal velocity

2mg

V- velocity p gas density A -frontal area Cd-drag coefficient

plugging in the values we will get ,

Mouse terminal velocity: 44.27 m/s,

Rat terminal velocity: 57.155 m/s.

b) consider the COM will travel half the distance of length

that is 2.5 cm and 10 cm for mouse and rat respectively

so, deacceleration

a= v²/2s

where s is the distance .

so, we get for mouse ,

a= 39196.658 m/s2

so, force = 0.030 *39196.658 m/s2 = 1175.88N

and for , rat we have,

a= 16333.47m/s2

so, force = 0.5 *16333.47m/s2 = 8166.73 N

c) area of mousebone =pi *d²/4 = 1.7671x10^-6 m²

so.load per unit area = 1175.88N/ 1.7671x10^-6 m²    = 6 .65 x10⁸ N/m2

area of rat bone = 7.068x10^-6 m²  

so.load per unit area = 8166.73 N/ 7.068x10^-6 m2 =1.15 x10⁹ N/m2

it seems both will get broken legs