Thursday

February 26, 2015

February 26, 2015

Posted by **rav** on Friday, July 14, 2006 at 5:23pm.

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 but i dont understand how to show the couple about the point with position vector i...but im assuming that it is a single force...but a couple cannot be put in equilibrium by a single force so it must be zero? is that correct???????

The unity direction vector for the forces are

(i+2j)/sqrt5 and

(I-2j)/sqrt5

F1= (3sqrt5)(i+2j)/sqrt5 )

F2= (sqrt5)(i-2j)/sqrt5

multiply those out, you get the cartesian coordinats in i,j

Ft= F1+F2=3i + 6j +i -6j=4i check that.

You have in the j direction components a couple (two equal forces, opposite directions)if the forces are acting at a distance. You do not state in the question where the forces act in relation to the center of mass position. I do not understand the comments you made on a couple.

(Broken Link Removed)

Do the "multiply those out," part for F2 again. You should also be aware of simple cancellation too. Take the root 5 out. (If I read the problem/answer right.)

Bob, it looks like responded to you and thought it was rav's post. Sorry about that.

rav, if you read this, yes, you treat the sum of 2 forces or vectors as a single force/vector. Vector addition is identical with Force addition. I also think your answer was correct.

your Ft is wrong...because F2 is equal to to i-2j...not i-6j...so Ft is therefore 4i+4j

The question i got was a

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 calculate the total force F[1] + F[2], acting on the particle

from there i was told to show that the couple of the total force about the point with position vector i is zero

this is all the information i was given

correct on the force ft=4i+4j.

Roger mention the equilbrium force, and on that he would be right. However, a couple is not an equilbrant, and I still dont understand how it applies here unless one knows the couple arm (distance from mass to the point the force is applied to). Thanks for checking my work. If you get more information, I am interested on the application of the "couple" here. For a force couple, the directions of the forces is not enough, it has to act on point remote from the center of rotation.

Ok rav, after looking at your question for the 3rd time, if you want it to be in equilibrium you need the vector to point in the other direction. multiply the sum force F you got by -1. Then it's in equilibrium.

**Answer this Question**

**Related Questions**

Calc - How close is the semi circle y= sqr.root of 16-x^2 to the point (1, sqr....

Algebra2/Trig - sqr(x^2-4)+(x^2)/x^2+1 I got ((x^2+1)*sqr(x^2-4)+x^2)/x^2+1 But ...

physics - Two forces of magnitude 20 newtons and 25 newtons respectively act at ...

Math - Find the quotient function f/g for f(x)=sqr(x+1) and g(x)= sqr( x-1). My...

Math - Find the quotient function f/g for f(x)=sqr(x+1) and g(x)= sqr( x-1). My ...

Algebra 2 - Ok, so I'm trying to find the inverse of each function. We're ...

physics - A 15.0-kg box is released from rest on a rough inclined plane tilted ...

math/physics - Forces with magnitudes of 2000 newtons and 900 newtons act on the...

vector and mass - A particle of mass m kg is acted on by two forces F1 and F2 ...

algebra - sqr root of 3a + 9(sqr root of 27a^3)