Cart Pushing Force

Problem Defininition

The goal of this analysis is to calculate the maximum pushing force that a cart can generate when is slowly pushes up against another object (we are neglecting impact forces). A hand writeup of the analysis shows the equations, while the images and videos below describe the physics in more detail.

When cart pushes 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.

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

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: r_in

The radius of Bracket 2 is: r_out

The radius of Wheel is: r_wheel

The input torque of the motor is τ_m

The output torque of the wheel is τ_out

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 τ_m, and the radius of Bracket 1 is r_in

so T = τ_m / r_in

Similarly, for the bracket 2, we know the radius of Bracket 2 is r_out and the output torque that we want to find is τ_out

so T = τ_out / r_out

Because T = T

so τ_m / r_in = τ_out / r_out

As a result, we can find τ_out in term of r_in, τ_m, andτ_m

τ_out = τ_{m *} (r_out / r_in)

Now we know that the output torque of the second bracket is τ_out . 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 τ_out = τ_{m *} (r_out / r_in).

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 = τ_out / r_wheel = (τ_{m *} (r_out / r_in))/r_wheel = τ_{m *} (r_out / (r_in* r _wheel ))

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 τ_out = τ_{m *} (r_out / r_in), 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: m_weight

_{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.

∑F_x = 0

_{ffriction -} F_push = 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 - m_car g - m_weight 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

m_car g* b + m_weight g* c - R₁* (a + b) = 0 .... Equation 3

Friction: ffriction = u* N

From equation 3, R₁ = (m_car g* b + m_weight g* c)/(a + b)

ffriction = u* R₁ = u* ((m_car g* b + m_weight g* c)/(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)