Conveyor belt and friction- Determine slip time

In summary, a series of small packages are moved by a conveyor belt over a 300mm radius pulley. The belt starts from rest and increases in speed at a constant rate of 150 mm/s^2. The coefficient of friction is 0.75. Using polar or path coordinates, the time the first package slips can be solved for by considering the sum of forces in the x and y direction and using the equations for polar coordinates. However, it can also be solved using cartesian coordinates by considering the maximum static friction and the equations for constant linear acceleration.
  • #1
JJBladester
Gold Member
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2

Homework Statement


A series of small packages is moved by a conveyor belt that passes

over a 300mm radius pulley. The belt starts from rest at t=0 and its

speed increases at a constant rate of 150 mm/s2. The

coefficient of friction is μ=0.75. Determine the time the first

package slips.

Answer: ts=11.32s

conveyor%20belt%20friction.jpg


Homework Equations


[tex]V_{0}=0[/tex]
[tex]r_{pulley}=300mm[/tex]
[tex]a_{pulley}=150mm/s^{2}[/tex]
[tex]\mu=0.75[/tex]

These are equations for polar coordinates, which I'm thinking I may need to use to solve the problem.
[tex]F_{r}=ma_{r}[/tex]
[tex]F_{\theta}=ma_{\theta}[/tex]
[tex]F_{net}=F_{r}{\bf e}_{r}+F_{\theta}{\bf e}_{\theta}[/tex]

[tex]m(\ddot{r}-r\dot{\theta}^{2})=F_{r}[/tex]
[tex]m(r\ddot{\theta}+2\dot{r}\dot{\theta})=F_{\theta}[/tex]

The Attempt at a Solution


First of all, I'm trying to decide if I should use cartesian, polar, or path coordinates for this problem.

If cartesian coordinates work, then I would set the problem up as below. Also, I don't think the radius of the pulley would matter...?

Sum of forces x-dir = -Fs+Fpulley=ma
Sum of forces y-dir = N-mg=0 -----> N=mg

The maximum static friction would be Fssmg.

For polar or path coordinates, I wouldn't know where to start.
 
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  • #2
I'm getting closer... I think.

I actually think I can solve this without using polar or path coordinates.

So, for constant linear acceleration, we have:

vf2=vi2+2a(x2-x1)

Because x2-x1 is really 2πr (arc length), we have:

vf2=vi2+2a(2πr)

I can solve this for vf and then plug that into vf=vi+at to get time... But I'm missing something with the whole Fssmg=ma... Like it should factor into the acceleration value for the two formulas above. Any help?
 

Related to Conveyor belt and friction- Determine slip time

1. What is a conveyor belt and how does it work?

A conveyor belt is a mechanical device used to transport materials from one location to another. It consists of a continuous moving belt that is powered by a motor, pulleys and rollers. The materials are placed on the belt and are carried along its length to the desired location.

2. What is friction in relation to conveyor belts?

Friction is the resistance force that occurs when two surfaces come into contact with each other. In the case of conveyor belts, friction is necessary to keep the materials on the belt and prevent them from slipping off.

3. How is slip time calculated for a conveyor belt?

Slip time is the time it takes for a material to slip from the conveyor belt. It is calculated by dividing the length of the belt by the speed of the belt. This can be expressed as: Slip time = Length of conveyor belt/ Speed of conveyor belt

4. How does friction affect slip time on a conveyor belt?

Friction plays a crucial role in determining slip time on a conveyor belt. Higher friction between the belt and the materials will result in a longer slip time, as the materials will have a stronger grip on the belt and will take longer to slip off.

5. What are some factors that can affect friction and slip time on a conveyor belt?

Some factors that can affect friction and slip time on a conveyor belt include the type and condition of the belt, the weight and surface texture of the materials being transported, and the speed of the belt. Other factors such as temperature, humidity, and the presence of lubricants or debris can also impact friction and slip time.

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