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How to find the mass of a conveyor belt so that it doesn't slip

  1. Jul 13, 2005 #1
    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]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?
  2. jcsd
  3. Jul 14, 2005 #2


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    I may be missing something, but it appears to me the final step in the solution may be incorrect. The 3108N appears to be the weight of the sand in the buckets on one side of the belt. The weight of the buckets and the belt is being ignored (or included as part of the hanging mass). I've never done the calculation, but apparently the exponential relating [tex] S_1 [/tex] and [tex] S_2 [/tex] comes from treating the normal force between the belt and roller distributed over the contact surface. I assume your book does that calculation. [tex] S_1 [/tex] and [tex] S_2 [/tex] appear to be the tensions in the belt at the upper roller. The larger tension includes half the weight of the hanging mass plus all of the weight of the sand, while the smaller tension is only half the hanging mass. The sum of those two tensions would be the force acting downward on the upper roller.

    It appears to me the mass attached to the lower roller should not include the mass of the sand that is adding to the tension on one side of the belt. Rather it should be twice the smaller tension, or 2*1984.

    It's possible I am misinterpreting what I am seeing, and I have not verified the exponential ratio of tensions, but it is clear that the first force is just the weight of the sand while the force at the end includes the weight of the mass hanging on the lower roller. Maybe you can sort it out if you take that difference into consideration.
  4. Jul 14, 2005 #3
    Hey thanks! Is it true that it does matter that the sand only affects one side of the belt, hence [itex]S_1 \neq S_2[/itex]? Because if sand was also present on the other side then the two would be equal right...?
  5. Jul 15, 2005 #4


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    Correct. If there were equal amounts of sand on both sides the tensions would be the same. The problem assumes you have the maximum imbalance of the tensions that can be achieved for the given coefficient of friction.
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