2D rigid body mechanics: force applied on sliding cart

[bilboDan]

Homework Statement

A cart of mass 20 kg is 1.8 m tall (along the y-axis) and 0.6 m long (along the x-axis). It sits on casters of negligible height which are located at the front and back edges of its length. A force of 100 N is applied from the left at a height of h from the ground to push the cart to its right in the positive x direction Assume that the casters are locked and that the coefficient of sliding friction is 0.25.
a. What is the acceleration of the cart?
b. What is the range of h such that the cart will not tip over? Show step by step calculations.

Relevant Equations

Newton's 2nd law; torque balance

The linear acceleration is easily found. The issue is finding the range of h over which the cart does NOT tip over. When this problem is submitted to [AI chatbot references redacted by the Mentors], the answer is h <= 0.588 m. This answer comes from balancing the clockwise torque about the front caster contact point from the applied force with the counterclockwise torque from gravity. However, this answer does NOT take into the account the acceleration of the cart, and the resultant torgue from the inertial force acting through the cart's center of mass. If we do take that into account, the answer is
h <= 1.047 m. Which is the correct answer?

 

[haruspex]

Homework Statement: A cart of mass 20 kg is 1.8 m tall (along the y-axis) and 0.6 m long (along the x-axis). It sits on casters of negligible height which are located at the front and back edges of its length. A force of 100 N is applied from the left at a height of h from the ground to push the cart to its right in the positive x direction Assume that the casters are locked and that the coefficient of sliding friction is 0.25.
a. What is the acceleration of the cart?
b. What is the range of h such that the cart will not tip over? Show step by step calculations.
Relevant Equations: Newton's 2nd law; torque balance

The linear acceleration is easily found. The issue is finding the range of h over which the cart does NOT tip over. When this problem is submitted to [AI chatbot references redacted by the Mentors], the answer is h <= 0.588 m. This answer comes from balancing the clockwise torque about the front caster contact point from the applied force with the counterclockwise torque from gravity. However, this answer does NOT take into the account the acceleration of the cart, and the resultant torgue from the inertial force acting through the cart's center of mass. If we do take that into account, the answer is
h <= 1.047 m. Which is the correct answer?

The least confusing approach is to work in an inertial frame and take moments about the mass centre of the cart. What does that give you?
You used a non-inertial frame, while the chatbots used an inertial frame and took moments about the caster. Having discovered which was right, the challenge is to figure out where the other went wrong.

 

[berkeman]

Thread is locked temporarily for Moderation...

 

[berkeman]

After redacting the reference to several AI chatbots, the thread is reopened provisionally. Thanks for your patience.

 

[bilboDan]

After redacting the reference to several AI chatbots, the thread is reopened provisionally. Thanks for your patience.

Many apologies for my references to AI chatbots. I had not read about the prohibition against such due to my carelessness. ( By the way, according to my understanding, none of the chatbots gave the correct answer, and I just wanted to confirm that.)

 

[bilboDan]

The least confusing approach is to work in an inertial frame and take moments about the mass centre of the cart. What does that give you?
You used a non-inertial frame, while the chatbots used an inertial frame and took moments about the caster. Having discovered which was right, the challenge is to figure out where the other went wrong.

No, I do not agree that it is correct to work in an inertial frame, i.e. the ground on which the cart is accelerating. Instead, one must work in the accelerated frame to calculate all torques. One must then include the inertial force associated with the cart's acceleration, and find that the answer for the range of h to be h<= 1,047 m.

 

[haruspex]

No, I do not agree that it is correct to work in an inertial frame, i.e. the ground on which the cart is accelerating. Instead, one must work in the accelerated frame to calculate all torques. One must then include the inertial force associated with the cart's acceleration, and find that the answer for the range of h to be h<= 1,047 m.

It is always valid to work in an inertial frame; no exceptions. And it gives the same answer.
It is also valid to work in any non-inertial frame you choose, provided you handle all fictitious forces correctly, as you have done.

So the question becomes, what did the AI do wrong?
If you take moments about a fixed point on the ground, you have to include the angular acceleration implied by the linear acceleration of the mass centre, $\vec a=\vec r\times\vec\alpha$.

 

[bilboDan]

I shall explain how the 2 different answers are obtained. First some definitions:
L = length of cart = 0.6m
H = height of cart = 1.8 m
m = mass of cart = 20 kg
F = applied force = 100 N
µ = coefficient of sliding friction = 0.25
and g = acceleration due to gravity = 9.8 ms-2

To obtain an answer of h<= 0.588 m, we say that
torque,clockwise = F h (torque about the front caster contact point)
torque,counter-clockwise = mgL/2 (torque about the front caster contact point)
For the cart not to tip over forwards, torque,clockwise <= torque,counter-clockwise
so h <= mgL/2F = 0.588 m

To obtain an answer of h <= 1.047 m
torque,clockwise = F h
torque,counter-clockwise = mgL/2 + (F - µmg)H/2
The term (F - µmg)H/2 is due to the inertial force acting in -x direction through the center of mass.
Then the condition torque,clockwise <= torque,counter-clockwise
leads to
h <= mgL/2F + (1 - µmg/F)H/2 = 1.047 m

I am reasonably confident that h <= 1.047 m is the correct answer.
My problem is that for me, the inertial force term only occurs as a fictitious force in a frame that is accelerating at 2.55 ms-2 , the frame in which the cart is stationary, In the inertial frame of the ground, we just have the torque due to the applied force and gravity, and that is all. If you say that it is possible to get the answer in the inertial frame of the ground, please explain how that can be done. Thank you.

 

[haruspex]

If you say that it is possible to get the answer in the inertial frame of the ground, please explain how that can be done.

Using the leading caster as the axis, the angular acceleration is $a/(H/2)$
Net torque about the leading caster $=Fh-mgL/2=ma/(H/2)(H/2)^2=maH/2$
$F-\mu mg=ma$
$Fh-mgL/2=(F-\mu mg)(H/2)$
$h=\frac{(F-\mu mg)H+mgL}{2F}=\frac{(100-0.25*20*9.8)1.8+20*9.8*0.6}{200}=1.047$.

 

[bilboDan]

This is an interesting approach I had not thought of. However, I am still having a problem.
If we are using the leading caster (at the front bottom corner of the cart) as the location of the axis about which torques and angular acceleration are calculated, the moment of inertia of the cart about that axis is :
I = m(L*L + H*H)/3
The relation of the angular acceleration of the center of mass about that axis to the horizontal acceleration a is given by:
α = a H/ (2*d*d) where d*d = ¼(L*L + H*H) = square of diagonal distance from
axis (at front caster) to c.m.
= 2aH/(L*L + H*H)
and equating the torque to the angular acceleration, we get:

Fh – mgL/2 = 2maH/3

This is not the same as what you showed:
Fh – mgL/2 = maH/2

I cannot see how I computed the moment of inertia or the angular acceleration incorrectly. Can you explain in more detail how you derived expressions for angular acceleration and moment of inertia about the front caster at the front bottom corner? Thanks.

 

[haruspex]

This is an interesting approach I had not thought of. However, I am still having a problem.
If we are using the leading caster (at the front bottom corner of the cart) as the location of the axis about which torques and angular acceleration are calculated, the moment of inertia of the cart about that axis is :
I = m(L*L + H*H)/3
The relation of the angular acceleration of the center of mass about that axis to the horizontal acceleration a is given by:
α = a H/ (2*d*d) where d*d = ¼(L*L + H*H) = square of diagonal distance from
axis (at front caster) to c.m.
= 2aH/(L*L + H*H)
and equating the torque to the angular acceleration, we get:

Fh – mgL/2 = 2maH/3

This is not the same as what you showed:
Fh – mgL/2 = maH/2

I cannot see how I computed the moment of inertia or the angular acceleration incorrectly. Can you explain in more detail how you derived expressions for angular acceleration and moment of inertia about the front caster at the front bottom corner? Thanks.

You don’t have to worry about the moment of inertia. The object is not actually rotating.

Consider a point mass m moving at velocity $\vec v$ at displacement $\vec r$ from a reference axis. Its angular momentum about the axis is $m\vec r\times \vec v$. If we integrate this over some body mass M that is not rotating then all parts have the same velocity vector and the aggregate angular momentum is $M\vec R\times \vec v$, where $\vec R$ is the displacement of the mass centre.

Differentiating, the rate of change of angular momentum is
$M\dot{\vec R}\times \vec v+M\vec R\times \dot{\vec v}$.
But $\dot{\vec R}=\vec v$, so this reduces to $M\vec R\times \dot{\vec v}$.
And rate of change of angular momentum about an axis equals the net torque about that axis.

It was probably misleading to refer to "angular acceleration" in post #9.

 

[bilboDan]

You don’t have to worry about the moment of inertia. The object is not actually rotating.

Consider a point mass m moving at velocity v→ at displacement r→ from a reference axis. Its angular momentum about the axis is mr→×v→. If we integrate this over some body mass M that is not rotating then all parts have the same velocity vector and the aggregate angular momentum is MR→×v→, where R→ is the displacement of the mass centre.

Differentiating, the rate of change of angular momentum is
MR→˙×v→+MR→×v→˙.
But R→˙=v→, so this reduces to MR→×v→˙.
And rate of change of angular momentum about an axis equals the net torque about that axis.

It was probably misleading to refer to "angular acceleration" in post

You don’t have to worry about the moment of inertia. The object is not actually rotating.

Consider a point mass m moving at velocity v→ at displacement r→ from a reference axis. Its angular momentum about the axis is mr→×v→. If we integrate this over some body mass M that is not rotating then all parts have the same velocity vector and the aggregate angular momentum is MR→×v→, where R→ is the displacement of the mass centre.

Differentiating, the rate of change of angular momentum is
MR→˙×v→+MR→×v→˙.
But R→˙=v→, so this reduces to MR→×v→˙.
And rate of change of angular momentum about an axis equals the net torque about that axis.

It was probably misleading to refer to "angular acceleration" in post #9.

 

[bilboDan]

You don’t have to worry about the moment of inertia. The object is not actually rotating.

Consider a point mass m moving at velocity $\vec v$ at displacement $\vec r$ from a reference axis. Its angular momentum about the axis is $m\vec r\times \vec v$. If we integrate this over some body mass M that is not rotating then all parts have the same velocity vector and the aggregate angular momentum is $M\vec R\times \vec v$, where $\vec R$ is the displacement of the mass centre.

Differentiating, the rate of change of angular momentum is
$M\dot{\vec R}\times \vec v+M\vec R\times \dot{\vec v}$.
But $\dot{\vec R}=\vec v$, so this reduces to $M\vec R\times \dot{\vec v}$.
And rate of change of angular momentum about an axis equals the net torque about that axis.

It was probably misleading to refer to "angular acceleration" in post #9.

I do not understand the derivation of the following equation (even though it finally gives kind of the right answer):

Fh – mgL/2 = (ma / (H/2)) (H/2)*(H/2) = maH/2

a.) The angular acceleration is calculated as "a/ (H/2)". That is only true if the
axis about which torques and angular acceleration are calculated is located at the bottom middle of the cart so that the lever arm is a distance of H/2 to the center of mass. But if the axis is at the bottom middle of the cart, the term mgL/2 on the left of the equation would not exist since the gravity force vector would have zero lever arm.

b.) It may be convenient to think that the cart body can be treated as a point mass m at the center of mass. However, total angular momentum does depend on the distribution of the mass, and so also the rate of change of total angular momentum.

As I said, I do not understand the reasoning behind this equation even though it does somehow finally give the right answer. More explanation would be appreciated.

 

[haruspex]

I do not understand the derivation of the following equation (even though it finally gives kind of the right answer):

Fh – mgL/2 = (ma / (H/2)) (H/2)*(H/2) = maH/2

a.) The angular acceleration is calculated as "a/ (H/2)". That is only true if the
axis about which torques and angular acceleration are calculated is located at the bottom middle of the cart so that the lever arm is a distance of H/2 to the center of mass. But if the axis is at the bottom middle of the cart, the term mgL/2 on the left of the equation would not exist since the gravity force vector would have zero lever arm.

b.) It may be convenient to think that the cart body can be treated as a point mass m at the center of mass. However, total angular momentum does depend on the distribution of the mass, and so also the rate of change of total angular momentum.

As I said, I do not understand the reasoning behind this equation even though it does somehow finally give the right answer. More explanation would be appreciated.

As I wrote in post #11, it was rather misleading for me to have described it as angular acceleration in post #9.
Hang on to the fact that net torque (about an axis) equals the rate of change of angular momentum about that axis.
The angular momentum of a body mass m moving irrotationally at velocity $\vec v$ wrt to the axis and displacement $\vec r$ from it is $m\vec r\times\vec v$. As I showed in post #11, the rate of change of that is $m\vec r\times\dot{\vec v}$. In the present case, with the leading caster as the axis, that is $m(H/2)a$.

If you are unhappy with this approach, you can take the mass centre as the axis instead, so angular momentum is constant 0. The torque balance is:
$F(h-H/2)+\mu mg(H/2)=mg(L/2)$
which gives the same result.

 

[haruspex]

It is interesting to see the relationship to the parallel axis theorem.

Consider an element $\delta m$ at displacement $\vec s$ from the mass centre of a body mass m moving at velocity $\vec v$ and rotating at rate $\vec \omega$.

The element has velocity $\vec v+\vec s\times\vec\omega$.

Its angular momentum about the origin is $(\vec r+\vec s)\times(\vec v+\vec s\times\vec\omega)\delta m=(\vec r\times\vec v+\vec s\times\vec v+\vec r\times(\vec s\times\vec\omega)+\vec s\times(\vec s\times\vec\omega))\delta m$.

By definition of mass centre, $\int\vec s.dm=0$, so the middle two terms vanish when we integrate. The first term reduces to $m\vec r\times\vec v$.

Expanding the last term by the triple vector product rule:
$\vec s\times(\vec s\times\vec\omega)=(\vec s\cdot\vec\omega)\vec s)-(\vec s\cdot\vec s)\vec\omega$.
But $\vec s$ and $\vec\omega$ are orthogonal, so their dot product is zero.

Meanwhile, $\int \vec s\cdot\vec s)\vec\omega.dm=\vec\omega\int s^2.dm=\vec\omega I$.

If the motion is a rotation about the origin then $\vec v=\vec r\times\vec \omega$, so $m\vec r\times\vec v=m\vec r\times(\vec r\times\vec \omega)=$
$m((\vec r\cdot\vec \omega)\vec r-(\vec r\cdot\vec r)\vec omega)=$
$-mr^2\omega$.

Fixing my sign errors I leave as an exercise to the reader.