- #1
Idioticsmartie
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1. Q: With what amount of force must a person pull on the vertical portion of the rope to make the 1000 kg box travel up the 45 degree rough ([tex]u_{k}[/tex] = 0.10) plane at constant speed?
Umm, here is a really awful diagram to help you get the picture. Mac doesn't come with a Paint program, so I grabbed a random one...and this is what I got. Again, sorry.
http://photos-a.ak.facebook.com/photos-ak-sctm/v123/24/75/1238100168/n1238100168_30084828_487.jpg
[tex]u_{k}[/tex] = 0.10
m = 1000 kg
[tex]\vartheta[/tex] = 45
g = 9.80 [tex]m/s^{2}[/tex]
[tex]\Sigma[/tex]F = m*a
[tex]F_{fr}[/tex]=[tex]u_{k}[/tex]*[tex]F_{n}[/tex]
Known Forces:
mg = 9800
[tex]F_{n}[/tex] = 6929.646
[tex]F_{fr}[/tex] = 692.9646
I think I did this problem correctly, but would someone mind terribly checking it? It's just that there aren't any examples in my book, so I want a second opinion before I turn this into my teacher. Thanks!
1) The co-ordinate plane is set so that the x-axis is the inclined plane and the y-axis is perpendicular (of course), so [tex]F_{n}[/tex] = y-axis.
[tex]F_{n}[/tex] is set perpendicular to the plane, while mg goes straight down. When you resolve mg, you get <6929.646, -6929.646>.
2) Therefore, [tex]F_{n}[/tex] = the y-component of mg, so [tex]F_{n}[/tex] = <0,6929.646>
3) [tex]F_{fr}[/tex] = [tex]u_{k}[/tex][tex]F_{n}[/tex] = 6929.646 * 0.10 = 692.9646
4)[tex] \Sigma[/tex] F = [tex]F_{a}[/tex]- [tex]F_{fr}[/tex]
[tex]F_{a}[/tex] > [tex]u_{k}[/tex] * [tex]F_{n}[/tex]
[tex]F_{a}[/tex] = 693.
Sigfigs, so it's 690 kg.
Is this correct?
Umm, here is a really awful diagram to help you get the picture. Mac doesn't come with a Paint program, so I grabbed a random one...and this is what I got. Again, sorry.
http://photos-a.ak.facebook.com/photos-ak-sctm/v123/24/75/1238100168/n1238100168_30084828_487.jpg
[tex]u_{k}[/tex] = 0.10
m = 1000 kg
[tex]\vartheta[/tex] = 45
g = 9.80 [tex]m/s^{2}[/tex]
Homework Equations
[tex]\Sigma[/tex]F = m*a
[tex]F_{fr}[/tex]=[tex]u_{k}[/tex]*[tex]F_{n}[/tex]
The Attempt at a Solution
Known Forces:
mg = 9800
[tex]F_{n}[/tex] = 6929.646
[tex]F_{fr}[/tex] = 692.9646
I think I did this problem correctly, but would someone mind terribly checking it? It's just that there aren't any examples in my book, so I want a second opinion before I turn this into my teacher. Thanks!
1) The co-ordinate plane is set so that the x-axis is the inclined plane and the y-axis is perpendicular (of course), so [tex]F_{n}[/tex] = y-axis.
[tex]F_{n}[/tex] is set perpendicular to the plane, while mg goes straight down. When you resolve mg, you get <6929.646, -6929.646>.
2) Therefore, [tex]F_{n}[/tex] = the y-component of mg, so [tex]F_{n}[/tex] = <0,6929.646>
3) [tex]F_{fr}[/tex] = [tex]u_{k}[/tex][tex]F_{n}[/tex] = 6929.646 * 0.10 = 692.9646
4)[tex] \Sigma[/tex] F = [tex]F_{a}[/tex]- [tex]F_{fr}[/tex]
[tex]F_{a}[/tex] > [tex]u_{k}[/tex] * [tex]F_{n}[/tex]
[tex]F_{a}[/tex] = 693.
Sigfigs, so it's 690 kg.
Is this correct?
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