Dynamics Question -- packages discharged from a conveyor belt

  • #1

Homework Statement


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

Homework Equations


F = ma

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.
 

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Answers and Replies

  • #2
PeroK
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When the package passes point A, what value for ##\theta## are you assuming?
 
  • #3
When the package passes point A, what value for ##\theta## are you assuming?
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
PeroK
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0 degrees as shown in the picture. But 90 degrees, counterclockwise from the (+) x axis (assuming the center of the circle is the origin).

So, what force is causing the acceleration in the x-direction?
 
  • #5
So, what force is causing the acceleration in the x-direction?
I assumed the conveyor belt was pushing horizontally on it, to offset frictional impacts.
 
  • #6
PeroK
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I assumed the conveyor belt was pushing horizontally on it, to offset frictional impacts.

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
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.
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
PeroK
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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?

I think so!
 
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