Calculating Net Force on an Object Moving in a Plane

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The discussion revolves around calculating the net force on an object moving in a plane, given its mass of 6.00 kg and position equations x = 4t^2 - 1 and y = 2t^3 + 6. Participants emphasize the need to derive the velocity and acceleration vectors from the position equations using calculus, specifically by finding the first and second derivatives with respect to time. Newton's second law, F = ma, is highlighted as essential for determining the net force. The conversation also addresses the importance of calculating instantaneous velocity rather than average velocity. Overall, the focus is on applying kinematic equations and derivatives to solve the problem effectively.
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I'm having terrible difficulty starting this problem, it's one of the chapter excercises in the book and it's revisited in the homework later on. I'm going to give different data, as I would like to actually solve this one myself, I just need a kick start...

Homework Statement


We've got an object moving in a plane, no velocity is stated at all, just that it's moving. It's mass is 6.00 kg and it's coordinates are given by 2 equations, x = 4t^2 - 1 and y = 2t^3 + 6. They are asking what the net force acting on this object is at time t = 5.00s.


Homework Equations


I know somewhere in there I'm going to use kinematic equations. I started by trying to find \Delta X and \Delta Y...



Thanks to anyone who can point me in the right direction!
 
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It's in a plane, so position, velocity, acceleration, and force are all vectors with x and y components. The position vector is (x,y)=(4t2-1,2t3+6). Can you find the velocity vector? (How is velocity related to position?) Then, can you find the acceleration vector? Then, can you find the force vector?
 
Ah, that makes a lot of sense! Thank you so much! I was completely overlooking that.
 
Glad to help.
 
If I'm using position to get a velocity vector with the formula V_x_{avg} = \frac{\Delta x}{\Delta t}, can I use t = 0 for my t_i?
 
You should be computing instantaneous velocity, not average velocity.

I assume this is a calculus-based course?
 
For the formulas I have you still need \Delta x and \Delta t

v_x = lim_{\Delta t \rightarrow 0}\frac{\Delta x}{\Delta t}

...and yes, this is calculus based.

I feel like I've missed a lesson or missed something in class.
 
That limit defines the derivative of x with respect to t. Given x as a simple function of t, say, x=t2, can you compute the derivative dx/dt ?
 
OH! So it would be 2t then... If the function was in fact t^2.
 
  • #10
crazy_shoes said:
I'm having terrible difficulty starting this problem, it's one of the chapter excercises in the book and it's revisited in the homework later on. I'm going to give different data, as I would like to actually solve this one myself, I just need a kick start...

Homework Statement


We've got an object moving in a plane, no velocity is stated at all, just that it's moving. It's mass is 6.00 kg and it's coordinates are given by 2 equations, x = 4t^2 - 1 and y = 2t^3 + 6. They are asking what the net force acting on this object is at time t = 5.00s.


Homework Equations


I know somewhere in there I'm going to use kinematic equations. I started by trying to find \Delta X and \Delta Y...

You would need two formulas:

1. Newton's second law \mathbf{F} = m \mathbf{a} and
2. definition of components of the acceleration vector

a_x = d^2x(t)/dt^2
a_y = d^2y(t)/dt^2

Eugene
 
  • #11
So, if my position in the x direction is a function of time, like x=2t^2 the derivative of that is 4t which should be my velocity in the x direction. Then a second derivative should give me 4 and that should be my acceleration in the x direction. Am I on the right track?
 
  • #12
crazy_shoes said:
So, if my position in the x direction is a function of time, like x=2t^2 the derivative of that is 4t which should be my velocity in the x direction. Then a second derivative should give me 4 and that should be my acceleration in the x direction. Am I on the right track?

Yes, you got it.

Eugene.
 
  • #13
Thanks! It's much appreciated. Good thing I have a whole week to finish studying for my test!
 
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