Forces with direction and magnitude

  • Thread starter dopey9
  • Start date
  • #1
32
0
Forces/couple!

particle of mass m kilograms is acted on by two forces F[1] and F[2] with magnitudes 3*sqr-root 5 newtons and sqr-root 5 newtons and directions parallel to the vectors i+2j and i-2j respectively.
The particle is initially at a position given by the vector 2i+j

i was told to calculate the cartesian components of F[1] and F[2] and hence calculate the total force F[1] + F[2], acting on the particle in component form

MY SOLUTION

The vector i+ 2j has length sqrt{1^2+ 2^2}= sqrt{5}[
A vector in that direction, with length is just 3 times that: 3i+ 6j. That's the first force vector.
Similarly, the vector i- 2j also has length root5 so that is the second force vector. The total force, then, is F1+ F2= (3i+ 6j)+ (i- 2j)= 4i+4j

from there i need to show the couple of the total force about the point with position vector i is zero.....so the total force is 4i+4j ............................

I would appreaciate a explanation to how to tackle this Q, so I can do it myself................but im assuming that it is a single force and a couple cannot be put in equilibrium by a single force so it must be zero? is that correct???????
 
Last edited:

Answers and Replies

  • #2
1,860
0
I'm not really sure where you are getting your calculations from. They might be right, but you would have to maybe explain them a little more because I don't really understand where you are getting things such as the length is just 3 times the slope of the force. The way I would approach the problem is to take your two forces and resolve them into component form, so {i_1; j_1; i_2; j_2}, and then add them together.
 
  • #3
nrqed
Science Advisor
Homework Helper
Gold Member
3,764
293
dopey9 said:
particle of mass m kilograms is acted on by two forces F[1] and F[2] with magnitudes 3*sqr-root 5 newtons and sqr-root 5 newtons and directions parallel to the vectors i+2j and i-2j respectively.
The particle is initially at a position given by the vector 2i+j

i was told to calculate the cartesian components of F[1] and F[2] and hence calculate the total force F[1] + F[2], acting on the particle in component form

MY SOLUTION

The vector i+ 2j has length sqrt{1^2+ 2^2}= sqrt{5}[
A vector in that direction, with length is just 3 times that: 3i+ 6j. That's the first force vector.
Similarly, the vector i- 2j also has length root5 so that is the second force vector. The total force, then, is F1+ F2= (3i+ 6j)+ (i- 2j)= 4i+4j
This all sounds correct to me
from there i need to show the couple of the total force about the point with position vector i is zero.....so the total force is 4i+4j ............................

This is where I am losing you. You said earlier that the particle is at the position 2i +j. So why are we talking about position "i" here?
And by
couple" what do you mean? The torque? I am not sure what 'couple" means.

I would appreaciate a explanation to how to tackle this Q, so I can do it myself................but im assuming that it is a single force and a couple cannot be put in equilibrium by a single force so it must be zero? is that correct???????
 
  • #4
32
0
..........

nrqed said:
This all sounds correct to me


This is where I am losing you. You said earlier that the particle is at the position 2i +j. So why are we talking about position "i" here?
And by
couple" what do you mean? The torque? I am not sure what 'couple" means.

but this is what i have taken from a past paper...it says "the couple of the total force about the point with position vector i is zero" ....its confusing me too...thankz for the help tho
 
  • #5
nrqed
Science Advisor
Homework Helper
Gold Member
3,764
293
dopey9 said:
but this is what i have taken from a past paper...it says "the couple of the total force about the point with position vector i is zero" ....its confusing me too...thankz for the help tho
Ok. By "past paper" do you mean it's a question taken from a book?!
Is there a drawing? Could you post the entire question as it is phrased there? Is this for an engineering class or a physics class? Have you covered torque?

I am asking all this simply because I wish I could be more helpful!

Patrick
 
  • #6
32
0
nrqed said:
Ok. By "past paper" do you mean it's a question taken from a book?!
Is there a drawing? Could you post the entire question as it is phrased there? Is this for an engineering class or a physics class? Have you covered torque?

I am asking all this simply because I wish I could be more helpful!

Patrick

Dont worry about it ...iv left it out..thankz for help...but if u do want the full question ...the first one i posted...and the last one i jus posted ...and thats all i was given..there was no diagram thankz anyway
 
  • #7
177
1
A couple is two equal and opposite forces times the perpindicular distance between them. Thus the body on which the couple acts will rotate, but not translate.
 
Last edited:
  • #8
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,551
586
dopey9 said:
particle of mass m kilograms is acted on by two forces F[1] and F[2] with magnitudes 3*sqr-root 5 newtons and sqr-root 5 newtons and directions parallel to the vectors i+2j and i-2j respectively.
The particle is initially at a position given by the vector 2i+j

i was told to calculate the cartesian components of F[1] and F[2] and hence calculate the total force F[1] + F[2], acting on the particle in component form

Ok, I'm ignoring everything in the solution, although some of it indicates you're answering another question

To break a vector into cartesian components, first you need to know the angle of the vector with respect to the x-axis. You would do this by taking arctan(y/x) of the direction it's in (y=j component, x=i component). Once you know the angle (call it a), you can find the x and y components by doing fcos(a) and fsin(a). Then add them together (keep in mind one force has positive and one negative y direction)
 

Related Threads on Forces with direction and magnitude

Replies
1
Views
484
Replies
12
Views
3K
  • Last Post
Replies
5
Views
2K
Replies
9
Views
3K
Replies
2
Views
1K
  • Last Post
Replies
13
Views
4K
Replies
1
Views
3K
  • Last Post
Replies
3
Views
4K
Replies
2
Views
5K
Replies
2
Views
828
Top