AH

Carnegie Mellon University

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
Problem 92
Problem 93
Problem 94
Problem 95
Problem 96
Problem 97
Problem 98
Problem 99
Problem 100
Problem 101
Problem 102
Problem 103
Problem 104
Problem 105

Problem 1

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.25 with the floor. If the train

is initially moving at a speed of 48 $\mathrm{km} / \mathrm{h}$ , in how short a distance

can the train be stopped at constant acceleration without causing

the crates to slide over the floor?

Answer

36$m$

You must be logged in to like a video.

You must be logged in to bookmark a video.

## Discussion

## Video Transcript

so here the greatest deceleration is provided by the maximum frictional force in this case, the normal forces equaling MG. And this is simply from the from the figure and the Newton. Using Newton's second law, we can say that the acceleration would be equaling the maximum static frictional force divided by and the Mass. And this would be equaling to the mat to the er coefficient of static friction multiplied by the normal force or, in this case, MG divided by EMS, the EMS cancel out. And this is giving us this coefficient of static friction multiplied by the acceleration due to gravity. And we can say that the shortest distance to stop would be that magnitude of this of the displacement in the ex direction. This would be equaling the velocity squared, divided by two times the acceleration. And in this case, it's me. The velocity squared, divided by two times the coefficient of static friction times G. And in this case we have a velocity in kilometers per hour. So we'll say 48 kilometers per hour, multiplied by 1000 meters for every one kilometer multiplied by one hour for every 3600 seconds. We're going to square this term. This will be divided by two multiplied by the coefficient of static friction of 20.25 multiplied by 9.8 meters per second squared. And we find that the magnitude of the displacement in the ex direction is going to be equaling two, approximately 36 meters. That is the end of the solution. Thank you for watching.

## Recommended Questions