How to Calculate Net Force Magnitude at Specific Time?

Click For Summary

Homework Help Overview

The problem involves calculating the net force acting on a 2.60 kilogram mass moving in a plane, with its position defined by specific equations for x and y coordinates over time. The context includes the use of the position update formula and the momentum principle.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss finding the velocity by taking derivatives of the position equations and question how to apply these results to find net force. There are attempts to clarify the use of the momentum principle and the position update formula.

Discussion Status

Some participants have indicated they have found a way to solve the problem but are exploring alternative methods that align with classroom restrictions. There is an ongoing discussion about the implications of using the momentum principle without the second law of motion.

Contextual Notes

Participants note that the teacher has not yet covered the use of the second law of motion (F=ma), which influences the methods they can use to approach the problem.

danielatha4
Messages
113
Reaction score
0

Homework Statement


A 2.60 kilogram mass is moving in a plane, with its x and y coordinates given by x = 5.40t2 - 1.30 and y = 2.75t3 + 1.80, where x and y are in meters and t is in seconds. Calculate the magnitude of the net force acting on this mass at t = 2.10 seconds.

Homework Equations


Our teacher has been making us use the position update formula and the momentum principle:
[tex]\Delta[/tex]r=Vavg[tex]\Delta[/tex]t
[tex]\Delta[/tex]P=Fnet[tex]\Delta[/tex]t

The Attempt at a Solution



I can figure out the velocity at a given point in both the y and x directions by finding the derivative:
dx/dt=8.4t
dy/dt=8.25t2

But I don't know how to apply this. I really don't know where to start.

Edit: sorry, dx/dt should be 10.8t
 
Last edited:
Physics news on Phys.org
danielatha4 said:

Homework Statement


A 2.60 kilogram mass is moving in a plane, with its x and y coordinates given by x = 5.40t2 - 1.30 and y = 2.75t3 + 1.80, where x and y are in meters and t is in seconds. Calculate the magnitude of the net force acting on this mass at t = 2.10 seconds.


Homework Equations


Our teacher has been making us use the position update formula and the momentum principle:
[tex]\Delta[/tex]r=Vavg[tex]\Delta[/tex]t
[tex]\Delta[/tex]P=Fnet[tex]\Delta[/tex]t



The Attempt at a Solution



I can figure out the velocity at a given point in both the y and x directions by finding the derivative:
dx/dt=8.4t
dy/dt=8.25t2

But I don't know how to apply this. I really don't know where to start.

Edit: sorry, dx/dt should be 10.8t

Well the second derivative of the position functions or the first derivative of the velocity functions should give you a(t). Then the second law F=ma. So you have then you should have the forces in the x and y direction at the time given. Add those two vectors together... pythag. theorem.
 
Thanks for the reply. I figured out that way of solving this problem after I posted it. However, our teacher won't let us use f=ma yet, as we haven't gotten to it in our class yet.

I no longer NEED help on this problem since I have the correct answer, but does anyone know how to do it with only the position update formula and momentum principle?
 
danielatha4 said:
I no longer NEED help on this problem since I have the correct answer, but does anyone know how to do it with only the position update formula and momentum principle?

Once you know the components of velocity, just use them to calculate the momentum at two different times, say [itex]t=t_0[/itex] and [itex]t=t_0+\Delta t[/itex], and substitute it into the momentum principle formula you gave. The smaller you make [itex]\Delta t[/itex], the more accurate your result will be; and in the limit that [itex]\Delta t\to 0[/itex], your result becomes exact (taking this limit is exactly the same thing as taking the derivative [itex]\frac{d\textbf{P}}{dt}[/itex], so this should be no surprise to you).
 

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
913
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
6K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K