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Cart Pushing Force

  • Problem Define 
When cart pushes blocks with heavy loads, sometimes the back wheel which is connected to the motor may slip. Sometimes, the motor may stall. In this experiment, we need to figure out what may cause the back wheel to slip or the motor to stall. 

This is a cart with a timing belt drive power system. 

 Cart 
 3D CAD  

Here is an interesting Youtube video covering the wheel slip and motor stall.

YouTube Video




In order to figure out the what may cause the cart wheel to slip or motor to stall, we need to separate this experiment into two parts. Frist, we need to figure out how the timing belt power train works. Second, we need to focus on the cart movement and find what may cause the cart wheel to slip or motor to stall.

Part 1: Timing Belt Power Train
(Click image to get detailed explanation on what is Timing Belt)

  • Introduction to Timing Belt (What is Timing Belt?)
A timing belt is a non-slipping mechanical derive belt. It is made as a flexible belt with teeth moulded onto its inner surface. It runs ove matching toothed pulleys. If we know the torque of the motor which is mounted into the first bracket, how many torque can be transmitted out to the second bracket?

From the above figure, the shaft of the motor is mounted into the center of bracket 1. Then, bracket 1 is connect to bracket 2 by using the timing belt. Last, bracket 2 shares the same rotating shaft with the wheel. 
  • Variables
The radius of Bracket 1 is: rin
The radius of Bracket 2 is: rout
The radius of Wheel is: rwheel
The input torque of the motor is τm
The output torque of the wheel is τout
Tension in the timing belt is: T
  • Assumption
    • Friction can be neglected 
    • Timing belt is perfectly matched
  • Analysis
Tension in the timing belt is always the same
so T = T
Force = Torque / distance
For the bracket 1, we know the input torque of the motor is τm, and the radius of Bracket 1 is rin
so T = τm rin
Similarly, for the bracket 2, we know the radius of Bracket 2 is rout and the output torque that we want to find is τout
so T = τout rout
Because T = T
so τrin τout rout
As a result, we can find τout  in term of rin, τm, andτm
τout  τm * (rout rin)
Now we know that the output torque of the second bracket is τout   . Because the output bracket and the wheel shares the same rotating shaft, the torque of the wheel is the same to the torque of the second bracket. As a result, the torque of the wheel is τout  τm * (rout rin). 
Last step, after we know the torque of the wheel, we can use Force = Torque / distance, to find what is the force corresponding to the torque. 
F = Torque / distance = τout / rwheel = (τm * (rout rin))/rwheel = τm * (rout / (rinwheel ))
Part 1 is done. Then we need to use the output torque of the wheel to figure out when the wheel will slip or when to motor will stall.

Part 2: Cart Movement
  • Introduction 
In this part, after we know the torque of the wheel is  τout  τm * (rout rin), we need to figure out when the wheel will slip and when the motor will stall.
  • Variables
Distance from the shaft of the back wheel to the cart's center of mass is: a 
Distance from the cart's center of mass to the shaft of the front wheel is: b
distance from the weight's center of mass to the shaft of the front wheel is: c
The mass of the cart is: mcart
The mass of the the weight is: mweight
The pushing force is Fpush
The coefficient of friction is u
The reaction force of the back wheel is R1
The reaction force of the front wheel is R2
The gravity is g
  • Assumption
    • Cart is at quasi-static motion 
    • The timing belt is perfectly matched 
    • There is no friction lost.
  • Analysis
Free Body Diagram: (FBD)

Sum of forces in the x (horizontal) direction should be 0. Pointing to the right is the positive direction.

 ∑Fx = 0

ffriction - Fpush = 0   ... Equation 1

Sum of forces in the y (vertical) direction should be 0. Pointing to the top is the positive direction.

 ∑Fy = 0

R1 + R2 - mcar g - mweight g = 0   ... Equation 2

Sum of the moment in the direction pointing out of the screen should be 0. Pointing out of the screen is the positive direction. Moment is force corss produce the arm. Check what is cross product.

 ∑Mo = 0

mcar gb + mweight gc - R1(a + b) = 0   .... Equation 3 

Friction:  ffriction = u* N

From equation 3, R= (mcar gb + mweight gc)/(a + b)

ffriction =  uR= u* ((mcar gb + mweight gc)/(a + b))

In order to figure out whether the wheel will slip or the motor will stall. We need to compare the friction force with the force transmitted out from the torque of the wheel.

If F > ffriction the wheel will slip.  (F is from part 1)

If F < ffriction the motor will stall. (F is from part 1)

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