Problem

Calculate the pressure due to the ocean at the bo…

05:26

Problem 81 Hard Difficulty

Suppose you hit a steel nail with a 0.500 -kg hammer, initially moving at 15.0 $\mathrm{m} / \mathrm{s}$ and brought to rest in 2.80 $\mathrm{mm}$ . (a) What average force is exerted on the nail? (b) How much is the nail compressed if it is 2.50 $\mathrm{mm}$ in diameter and 6.00 -cm long? (c) What pressure is created on the $1.00-\mathrm{mm}-$ diameter tip of the nail?

Answer

(a) $F=40200 \mathrm{N}$
(b) $\Delta L=2.34 \cdot 10^{-3} \mathrm{m}$
(c) $p=5.12 \cdot 10^{10} \mathrm{N} / \mathrm{m}^{2}$

Related Courses

Physics 101 Mechanics

College Physics for AP® Courses

Chapter 11

Fluid Statics

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Christina K.

Rutgers, The State University of New Jersey

Watch More Solved Questions in Chapter 11

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Video Transcript

We're told that we hit a steel now with 0.5 kilogram hammer that's initially moving at 15 meters per second and brought the rest in 2.8 millimeters. So first we're asked. But the average force is exerted on the nail. So I've drawn a little diagram. Here we have a hammer coming down at 15 meters per second. It drives the nail into the wood or whatever it is, and at a distance 2.8 millimeters. And at this point the hammer is at rest. So we want to figure out what the average forces to do that we need to figure out what the average, um acceleration is. Our deceleration is so we have our values here and our average acceleration is the change in velocity divided by the change in time. Well, we can rewrite that. It's the change in velocity divided by the change in displacement multiplied by the change in displacement provided by the change in time. So we have the change in velocity. In the end, we get change in velocity squared, divided by the change in displacement, the distance so we can then values in and we get that rather large acceleration of 88,400 meters per second squared, which gives us if the hammer weighs half a kilogram. That gives us a forced average force of 40.2 killers. Now ask how much is the nail compressed if it is 2.5 millimeters in diameter and six centimeters long so we can look at our equations. We have the cross sectional area of the nail that we could get by knowing its diameter. And the Youngs module is of steel is 210 Guica Pascal's The stress than in the nail is the force that we had before times four, divided by pi times the diameter squared. And if we plug those numbers and we get 8.2 Giga Pascal's now we know that the strain is just The stress divided by the Youngs module is, and there's also the change in length divided by the initial. So our change in length is the stress divided by the Youngs module ist times the initial length Klink and plug numbers in for that, and we get that the nail is compressed or changes its length by 0.23 Senate years just fight a lot. So that's just while the hammer is acting on it. And hopefully that now doesn't buckle. Or we would cover back now and finally were asked what the pressure is created on the one millimeter diameter tip of the We have a sharp point on the nail so we can find if we know the every force and we have the area of the tip. We confined the pressure, so that is our average force kinds four over pi kinds. The one millimeter diameter one is 10 times my 10 to the minus three square, and we get that that is 51 point to take a pascal's.

We have video lessons for 95.57% of the questions in this textbook