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Im having problems doing this example question, the answeres are given under each question but I dont understand how to get there...

I had to upload this question as an image file because there is a diagram to this question.
http://img76.imageshack.us/img76/7307/staticsir4.jpg

Thanks in advance!

2007-05-29 13:10:25 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

Center of mass

The bar center of mass is 2.87/2 * 8.9 at an angle of 50

for x:
Cos(50)*2.87/2 * 8.9
8.21
y:
sin(50)*2.87/2*8.9
9.78

the hanging crate
x=cos(50)*2.15*30.4
42.0
y=(sin(50)*2.15-1.1)*30.4
16.63

the resultants
x:
(8.21+42)
50.21
or 1.28 m
y:
(9.78+16.63)
26.41
or
.672 m



To calculate F, sum the torques about the pivot that are perpendicular to the rod:
0=9.81*cos(50)*
(2.87/2*8.9+2.15*30.4)-
2.87*sin(50)*F

F=(9.81*cos(50)*(2.87/2*8.9+2.15*30.4))/(2.87*sin(50))

F=224 N

The upward reaction force is the sum of the downward:
9.81*(30.4+8.9)
=385 N

The horizontal is equivalent to the F
224
The resultant sqrt(224^2+385^2)
445 N
angle
Atan(385/224)
59.8 degrees

With friction, the normal force times µ must be greater than the horizontal force
so 385*µ>224
the minimum µ is
224/385
0.582


j

2007-05-30 05:31:53 · answer #1 · answered by odu83 7 · 0 0

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