# Modeling Quickly Pushed Boxes: A Beginner's Guide

• B
• walking
In summary: I am having trouble understanding how the impulse affects the motion.In summary, it seems that you are having difficulty understanding how the impulse affects the motion.
walking
I am a beginner to mechanics and am finding it very difficult to visualize/model what happens if I push a box quickly and let go. I know how to model the scenario when the box is continually pushed by a force. We simply use ##F=ma## where ##F## is the force. If the ground is rough, we also use the equation for static friction: if ##F<f_s## then the box doesn't move. If ##F>f_s## then the box moves and then there will be a kinetic friction ##f_k## afterwards.

However, I am not sure why I am finding it difficult to model the situation when the box is only pushed once. I think it has something to do with impulses, eg if someone is running in one direction and someone pushes them to the right then their directoin will change slightly. However I don't know the maths behind this in order to be able to model it. I have a feeling it is really simple but I am just not seeing it.

A similar scenario would be someone sliding on ice and being pushed momentarily backwards. How do his suvat variables change? Let's say his initial speed was ##v_0## and his acceleration was ##-f_k## (kinetic friction). What happens when someone momentarily (rather than continuously) pushes him backwards to try to make him stop? I don't know why but I am finding this difficult to model.

Any help would be appreciated.

After the box loses contact with your hands, the relevant equation is still ##F=ma## with the initial condition that the box has an initial velocity. The maths are the same.

kuruman said:
After the box loses contact with your hands, the relevant equation is still ##F=ma## with the initial condition that the box has an initial velocity. The maths are the same.
Could you please give an example? I'm not sure i fully understand yet. I think my main problem is modelling the relationship between the push force and the "initial velocity" you mentioned. Is there an equation relating my push force ##F## to the subsequent velocity of the box?

walking said:
We simply use ##F=ma## where ##F## is the force.
F is the vector sum of all forces here.

walking said:
Is there an equation relating my push force ##F## to the subsequent velocity of the box?
The forces give you the acceleration. Do you understand the relation between acceleration and velocity?

Place a book on the floor and give it a push with your toe. Watch it slide across the floor until it comes to a stop. There are two stages to the mathematical analysis of the motion:
1. While your toe is in contact with the book the equation governing the motion is
##m\frac{dv}{dt}=F_{toe}-F_{friction}##
At this stage the force that the toe exerts on the book is greater than the force of friction and the book picks up speed as it accelerates forward. At the end of this stage, you stop pushing with your toe and the book loses contact with the toe which marks the beginning of the second stage.
2. The book has lost contact with the toe but has initial speed ##v_0## the speed it acquired by the end of the previous stage. The equation governing the motion is
##m\frac{dv}{dt}=-F_{friction}##

anorlunda and russ_watters
I think, if you have a complex situation, you need to break it down into simple parts, which are each manageable.

You mention SUVAT equations: these apply to constant acceleration situations. So if forces change, the acceleration may change and SUVAT equations don't work. But you can still use them for the sections of constant acceleration separately.
Simple vehicle questions are often like this: there is a period of acceleration, a period of constant velocity and a period of deceleration.

You mention impulse: in that case you may not know what the forces and accelerations are during the impulse, but you do know the change in momentum and hence the change in velocity.
Calculate the motion up to the collision. Calculate the change in velocity due to the impulse. Then calculate the motion after the impulse.

A.T. said:
x
kuruman said:
x
Merlin3189 said:
x
I think my problem is exactly how the force affects the velocity. It seems intuitive that if I know the mass of something, the coefficient of kinetic friction and the applied force, I should be able to figure out the speed.

I'll try to put it into a question. Let's say a skater of mass ##80kg## is sliding on ice at velocity ##5ms^{-1}##. Also ##\mu_k=0.5##.
a) How long does it take the man to come to a stop if he doesn't do anything?
b) WHat about if he applies a momentary force of ##10N## with his right foot on the ice in the forward direction, so that the ice applies ##10N## backwards on him. After that he doesn't do anything.

Sorry if I am overcomplicating this, I am just not seeing it at the moment.

walking said:
I think my problem is exactly how the force affects the velocity.
Again, do you understand the relation between acceleration and velocity?

russ_watters
walking said:
I think my problem is exactly how the force affects the velocity.

Do you understand the equations
##v=v_0+at## and ##F_{net}=ma##?
If yes, what do you get if you use the second equation to eliminate the acceleration in the first equation?

walking said:
I think my problem is exactly how the force affects the velocity.
It sounds kind of like you think velocity is a function of force; it isn't.
b) WHat about if he applies a momentary force of ##10N## with his right foot on the ice in the forward direction, so that the ice applies ##10N## backwards on him.
How long is a moment?

...and how can ice provide a force? Isn't it frictionless?

walking said:
I am a beginner to mechanics and am finding it very difficult to visualize/model what happens if I push a box quickly and let go. I know how to model the scenario when the box is continually pushed by a force. We simply use ##F=ma## where ##F## is the force. If the ground is rough, we also use the equation for static friction: if ##F<f_s## then the box doesn't move. If ##F>f_s## then the box moves and then there will be a kinetic friction ##f_k## afterwards.

However, I am not sure why I am finding it difficult to model the situation when the box is only pushed once. I think it has something to do with impulses, eg if someone is running in one direction and someone pushes them to the right then their directoin will change slightly. However I don't know the maths behind this in order to be able to model it. I have a feeling it is really simple but I am just not seeing it.

A similar scenario would be someone sliding on ice and being pushed momentarily backwards. How do his suvat variables change? Let's say his initial speed was ##v_0## and his acceleration was ##-f_k## (kinetic friction). What happens when someone momentarily (rather than continuously) pushes him backwards to try to make him stop? I don't know why but I am finding this difficult to model.

Any help would be appreciated.
Simply continue utilizing Newtons law. (F) = (M) × (A) 1.Force 2.Mass 3.Acceleration With the 'Mass (Weight)' and the Acceleration (Speed & **Velocity) the Force is correct after multiplying M × A =F

Science-IS
walking said:
Is there an equation relating my push force FFF to the subsequent velocity of the box?
The effect of a Force cannot be determined on its own. Important extra information that is needed: the Time for which the force acts (Impulse= Change in Momentum = Force time time applied) or the Distance that the Force acts over (Added Kinetic Energy = Work done = Force times distance over which the force acts).
You can use either to work out the overall change in velocity.

Gregory Alan said:
Simply continue utilizing Newtons law. (F) = (M) × (A) 1.Force 2.Mass 3.Acceleration With the 'Mass (Weight)' and the Acceleration (Speed & **Velocity) the Force is correct after multiplying M × A =F

Science-IS said:

No It's not 100% correct. I misunderstood the initial differential expressions. Unfortunately I didn't see the threads above with the values.

F = d (m × v) / dt

F = m × dv/dt
magnitude, direction have to be recognized, the (v) velocity, (a) acceleration, (p) momentum, (f) force are vector quantities.

Newtons second law defines force to be equal to change in (p) momentum with a change in time. Momentum is defined to be the mass of an object times its velocity.
p = mv [m × v = p]
p = momentum (kg·m/s).
m = mass (kg)
v = velocity (m/s)

This differential equation can be solved with the boundary conditions that we described above if the variation is known with (f) force as a function of time.

Let us assume that the mass remains a constant value equal to (m). This assumption is pretty good for an airplane, the only change in mass would be for the fuel burned between point "1" and point "0". The weight of the fuel is probably small relative to the weight of the rest of the airplane, especially if we only look at small changes in time.. If we were discussing the flight of a baseball, then certainly the mass remains a constant. But if we were discussing the flight of a bottle rocket, then the mass does not remain a constant and we would have to specify how the mass varies with time to perform the integration. The Law is worked as (m) constant vice versa and in this form

F = m × dv / dt

The derivative of Velocity with respect to Time is the definition of (a) acceleration. The second law is then reduced to the familiar values of mass and an acceleration:

F = m × a

This relation divergence is only good for objects that have a constant mass. This equation tells us that an object subjected to an external force will accelerate and that the amount of the acceleration is proportional to the size of the force. The amount of acceleration is also inversely proportional to the mass of the object; for equal forces, a heavier object will experience less acceleration than a lighter object. Considering the momentum equation, a force causes a change in velocity; and likewise, a change in velocity generates a force. The equation works both ways.

The velocity, force, acceleration, and momentum have both magnitude and a direction associated with them. Scientists and mathematicians call this a vector quantity. The equations shown here are actually vector equations and can be applied in each of the of the component directions. We have only looked at one direction, and, in general, an object moves in all three directions (up-down, left-right, forward-back).

The motion of a craft resulting from aerodynamic forces the crafts mass/weight and thrust can be computed by using the second law of motion. In the example a fundamental problem arises when dealing with aerodynamic forces. The aerodynamic forces depend on the square of the velocity. As I said integrating the differential equations becomes a bit more complicated, but it is intrinsically trickier but nonetheless quantifiable.

We started with pushing a box and now we are talking about rockets and aerodynamic drag. Let's go back to pushing a box. @walking , can you solve the following problem using ##F_{net}=ma## without using work-energy considerations?

A box of mass ##m## is held at the end of a spring of spring constant ##k##. The spring is compressed from its relaxed position by displacement ##d##. The mass is on a rough table top such that the coefficient of kinetic friction between the box and the table is ##\mu_k.## The mass is released, slides across the table and eventually stops. Find the total distance traveled by the mass from the point of release to the stopping point.

Model: The force exerted by a spring is ##F_{spring}=-kx## where ##x## is the compression/extension of the spring from its relaxed position; the force of kinetic friction is ##f_k=\mu_k mg##.

kuruman said:
We started with pushing a box and now we are talking about rockets and aerodynamic drag. Let's go back to pushing a box. @walking , can you solve the following problem using ##F_{net}=ma## without using work-energy considerations?

A box of mass ##m## is held at the end of a spring of spring constant ##k##. The spring is compressed from its relaxed position by displacement ##d##. The mass is on a rough table top such that the coefficient of kinetic friction between the box and the table is ##\mu_k.## The mass is released, slides across the table and eventually stops. Find the total distance traveled by the mass from the point of release to the stopping point.

Model: The force exerted by a spring is ##F_{spring}=-kx## where ##x## is the compression/extension of the spring from its relaxed position; the force of kinetic friction is ##f_k=\mu_k mg##.
Yeah I was a bit late, I didn't see all the posts until late today. I was using this particular example for Science IS.

## 1. What is the purpose of "Modeling Quickly Pushed Boxes: A Beginner's Guide"?

The purpose of this guide is to provide beginners with a step-by-step tutorial on how to model quickly pushed boxes in a scientific context. This technique is commonly used in physics and engineering to simulate the motion of objects under external forces.

## 2. Do I need any prior knowledge or experience to use this guide?

No, this guide is designed for beginners with little to no prior knowledge in modeling or physics. However, a basic understanding of mathematical concepts and computer software may be helpful.

## 3. What software or tools are needed to model quickly pushed boxes?

This guide uses a free and open-source software called Blender, which is available for download on their official website. Other tools such as a computer mouse and keyboard may also be necessary for navigating the software.

## 4. Are there any limitations to modeling quickly pushed boxes?

Yes, this technique is a simplified version of real-world physics and may not accurately represent all scenarios. It also assumes certain conditions, such as a frictionless surface, which may not be present in real-life situations.

## 5. Can I apply this technique to other types of objects besides boxes?

Yes, this technique can be applied to any object that can be represented as a rigid body. However, the shape and properties of the object may affect the complexity of the modeling process.

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