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TSN79
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I postet this problem earlier without really getting anywhere so now I'll post the whole thing:
A vertical conveyorbelt has two rollers. A mass hangs in the center of the bottom one to make it taut. On this this belt there are small buckets that scoop up sand in the bottom and then empty this once they turn on top. The center distance between the rollers is 10 m, and there is a bucket every 0,5 m. Each bucket takes 20 dm3 of sand and is filled 80%. The sand has a density of 1800 kg/m3. There is also a friction coefficient between belt and rollers of 0,3.
I'm supposed to find what this mass at the bottom should be to prevet the belt from slipping. The calculation is done as follows:
[tex]Force=20 \cdot 1,8 \cdot 11 \cdot 0,8 \cdot 9,81=3108 N[/tex]
[tex]\frac{S_1}{S_2}=e^{\mu \alpha} \Rightarrow S_1=S_2 \cdot e^{0,3 \cdot \pi}=2,57 \cdot S_2[/tex]
[tex]F=S_1 - S_2 = 2,57 \cdot S_2 - S_2=1,57 \cdot S_2[/tex]
[tex]F=1,57 \cdot S_2 \Rightarrow S_2=\frac{3108}{1,57}=1984N[/tex]
[tex]S_1=S_2 \cdot 2,57=5092N[/tex]
[tex]F_a=S_1+S_2=5092+1984=7076N[/tex]
[tex]m=\frac{7076}{9,81}=721 kg[/tex]
These aren't my calculations, it's in the book, but I have a question:
What is the difference between the force calculated in the beginning (3108N) and the force found to be 7076N? Can someone please help me out here?
A vertical conveyorbelt has two rollers. A mass hangs in the center of the bottom one to make it taut. On this this belt there are small buckets that scoop up sand in the bottom and then empty this once they turn on top. The center distance between the rollers is 10 m, and there is a bucket every 0,5 m. Each bucket takes 20 dm3 of sand and is filled 80%. The sand has a density of 1800 kg/m3. There is also a friction coefficient between belt and rollers of 0,3.
I'm supposed to find what this mass at the bottom should be to prevet the belt from slipping. The calculation is done as follows:
[tex]Force=20 \cdot 1,8 \cdot 11 \cdot 0,8 \cdot 9,81=3108 N[/tex]
[tex]\frac{S_1}{S_2}=e^{\mu \alpha} \Rightarrow S_1=S_2 \cdot e^{0,3 \cdot \pi}=2,57 \cdot S_2[/tex]
[tex]F=S_1 - S_2 = 2,57 \cdot S_2 - S_2=1,57 \cdot S_2[/tex]
[tex]F=1,57 \cdot S_2 \Rightarrow S_2=\frac{3108}{1,57}=1984N[/tex]
[tex]S_1=S_2 \cdot 2,57=5092N[/tex]
[tex]F_a=S_1+S_2=5092+1984=7076N[/tex]
[tex]m=\frac{7076}{9,81}=721 kg[/tex]
These aren't my calculations, it's in the book, but I have a question:
What is the difference between the force calculated in the beginning (3108N) and the force found to be 7076N? Can someone please help me out here?