Problem

$\bullet$$\bullet$ A crate of 45.0 -kg tools rest…

09:06

Problem 48 Hard Difficulty

$\bullet$$\bullet$ A person pushes on a
stationary 125 N box with
75 $\mathrm{N}$ at $30^{\circ}$ below the hori-
zontal, as shown in Fig-
ure 5.56 . The coefficient of
static friction between the
box and the horizontal floor
is 0.80 . (a) Make a free-
body diagram of the box. (b) What is the normal force on the
box? (c) What is the friction force on the box? (d) What is the
largest the friction force could be? (e) The person now replaces the
his push with a 75 $\mathrm{N}$ pull at $30^{\circ}$ above the horizontal. Find the
normal force on the box in this case.

Answer

a) 4.05 $\mathrm{m} / \mathrm{s}^{2}$
b) 162.5 $\mathrm{N}$
c) 130 $\mathrm{N}$
d) 160 $\mathrm{N}$
e) 87.5 $\mathrm{N}$

Related Courses

Physics 101 Mechanics

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Watch More Solved Questions in Chapter 5

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

Problem 88

Problem 89

Problem 90

Problem 91

Video Transcript

the free body diagram for this part is shown here. No force acting upwards, crabbed, inciting downwards. And then you have the applied force the Porsche, which we're calling p exacting at an angle 30 degrees from the horizontal. So this is the angle and and said And so you have a fictional force just opposing the motion of defecting to the right. So to start party, it's decompose thiss pushing for us and do its vertical and horizontal components. Say you have do any degree angle here and this is P. So the horizontal component she's hacking in the negative wax exact system. This thinking of extraction, this case is P CO sign 30 and the vertical component the negative. Why Axis is P sign 30. Okay, so this is to set up and what this means is out. We can resolve forces in the white direction, which will be zero. This would lead to end in the up upward for a put force. The core energy, uh, gravity, plus the white component P signed 30 like a part of the pushing force on dso MGs. Just 1 25 1 25 your months and pia 75 And so we get normal for us of 1 63 news. Okay, Excellent on part. See, Similarly similar to part B. The result forces the extraction. And and again, there's no there's no net force. And so we have f ah, which is a frictional forces. He called to Pete in the horizontal component pushing force of Pecos 30. And so this s o. This is 75 coast 30. So are you frictional Force 65 needs party. Um, you have a party. You have no, um, use of ass that is given and we know end. So you're frictional force. Your fictional force will be well, some simply be are your static friction force will simply be mischievous times and the normal force. And so mischievous is 0.8 normal force this 1 63 as we found in part B. So this comes out to one thirteen 13 and finally part E In partly, you have the same everybody diagram, but one adjustment, which is r. Well, I guess it would be a couple of adjustments here. But the main difference here is that this is a pulling force now P is acting. Ah is acting to the diagonal, right? And so the affection would be to the left, too. So in this case, you still in the still of the same angle on everything but the vertical component of pea Eye's acting up words this time. And the horizontal compartment P is acting to the right this time. So, um so this is Sparky. So we resolve forces in the white direction. And again, that is, there is no Net Force, sir. And so you have n plus p sign 30 This time equals and she and so and is just em xi minus pew. Sign 30. So that's 1 25 minus 75 70 30 which gives us 88 newts in this case.

Get More Help with this Textbook