# Dynamics Question -- packages discharged from a conveyor belt

1. Feb 5, 2017

### RoyalFlush100

1. The problem statement, all variables and given/known data
A series of small packages, each with a mass of 4 kg, are discharged from a conveyor belt as shown. Assume that the coefficient of static friction between each package and the conveyor belt is 0.4.

Determine the force exerted by the belt on a package just after it has passed Point A

I attached the picture below

2. Relevant equations
F = ma

3. The attempt at a solution
So I began by drawing a FBD:
-W: Weight pointed downwards
-N: Normal pointed upwards
-f: Friction pointed left
-B: Force from the belt pointed right.
From what I understand, we are trying to solve for B.
Also, it is about to enter a circular region, which makes me think that the net acceleration is -mv^2/r in the y direction and dv/dt in the x direction.

So, I began by solving the forces in the y direction:
-mv^2/r = -W + N
-(4)(1^2)/(0.250) = -(4)(9.81) + N
--> N = 23.24 N

From there, I setup the equation in the x direction:
dv/dt = B - (u)N
dv/dt = B - (0.4)(23.24)
Now the thing is, I assumed that the speed is consistent which would imply B = (0.4)(23.24) = 9.296 N.

However, this answer was wrong, so I am not sure how to approach it from here.

#### Attached Files:

• ###### problem 1.jpeg
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Last edited by a moderator: Feb 5, 2017
2. Feb 5, 2017

### PeroK

When the package passes point A, what value for $\theta$ are you assuming?

3. Feb 5, 2017

### RoyalFlush100

0 degrees as shown in the picture. But 90 degrees, counterclockwise from the (+) x axis (assuming the center of the circle is the origin).

4. Feb 5, 2017

### PeroK

So, what force is causing the acceleration in the x-direction?

5. Feb 5, 2017

### RoyalFlush100

I assumed the conveyor belt was pushing horizontally on it, to offset frictional impacts.

6. Feb 5, 2017

### PeroK

Friction is a resisting force. Unless an external force is trying to accelerate an object, the frictional force will be zero. In the case of a block at rest or a block moving with constant velocity horizontally, there can be no frictional force; if there were the block would accelerate.

7. Feb 5, 2017

### RoyalFlush100

Okay, so then we're left with its weight and the normal force. So would the normal force be defined as the "the force exerted by the belt on a package." In which case the answer would simply be 23.24 N?

8. Feb 5, 2017

I think so!