Sign up for our free STEM online summer camps starting June 1st!View Summer Courses

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

Need more help? Fill out this quick form to get professional live tutoring.

Get live tutoring
Problem 32

A block is pushed across a floor by a constant force that is applied at downward angle $\theta($ Fig. $6-19) .$ Figure $6-36$ gives the acceleration magnitude $a$ versus a range of values for the coefficient of

kinetic friction $\mu_{k}$ between block and floor: $a_{1}=3.0 \mathrm{m} / \mathrm{s}^{2}, \mu_{k 2}=$

$0.20,$ and $\mu_{k 3}=0.40 .$ What is the value of $\theta ?$

Answer

$\theta=60^{0}$

You must be logged in to like a video.

You must be logged in to bookmark a video.

## Discussion

## Video Transcript

So let's draw first the free body diagram for the system. We have a point. Mass will treat it as a point ness. Uh, going up is forced. Normal. Going down is you wait mg at an angle. We have f making an angle with the horizontal Seita and opposite to the direction of motion would be the force of fiction. At this point, we can apply the sum of forces in the ex direction. This will equal the mass times acceleration. This would equal f co sign of data minus the force of friction. We can say that some of forces in the UAE direction would be equal to forced normal minus the force sign if ADA minus M. G, this is equaling zero because there isn't any acceleration in the UAE direction we could we know that here the force of friction would be equal to the coefficient of kinetic three times the force normal and according to this equation, this will be equal to the cowfish for of the coefficient of friction, the kinetic coefficient of friction multiplied by M G plus F sign off data. So at this point, we can then substitute this into the first equation and say f co sign of Fada minus the coefficient. Kinetic friction times M G minus f Sign of data would be equal to the mass times acceleration and so we can solve for the acceleration. This is equaling F over m co sign of Fada minus the coefficient of kinetic friction times Sign of data. This is equaling rather minus the coefficient of friction times g from the figure we know that a is equaling 3.0 ah, 3.0 meters per second squared and we know that the coefficient of kinetic friction is equal to zero. So this essentially means that 3.0 meters per second squared, it would be equal to f co sign of data divided by M Um, we also know that at a equaling zero meters per second and the coefficient of kinetic friction equaling 0.20 we can say that zero equals f over and Times co sign of theta minus 0.2. Sign of fada. Close parentheses minus, Uh, point. Okay, let's through type. That's given your workbook from these parameters. Here we can say that zero equals f over M Times Co sign of data minus 0.20 Sign of Fada closed parentheses minus 0.20 times 9.8 meters per second squared. And so at this point, we can then say that this F over m co sign a fada minus 0.20 sign of Fada close parentheses minus 0.20 times 9.80 meters per second squared is gonna be equal to 3.0 meters per second squared minus 0.20 I've been saying minus 0.20 f over M sign of fada minus 1.96 meters per second squared. We condone say that zero is gonna be equal to 1.4 meters per second squared minus 0.20 times f sign of Fada divided by M and we can then say 5.2 zero meters per second squared equals f sign of Fada over m and we're going to combine the two results combining this with this equation. We find that tangent of Fada will equal 5.2 meters per second squared, divided by 3.0 meters per second squared. This is equaling 1.73 fatal with n equal our time of 1.73 This is giving us approximately 60 degrees. So this is our angle theta. That would be our final answer. That is the end of the solution. Thank you for watching.

## Recommended Questions