Two forces vecF_1=hati+5hatj and vecF_2=3hati-2hatj act at points with two position vectors respectively hati and -3hati +14hatj How will you find out the position vector of the point at which the forces meet?

2 Answers
Jun 16, 2016

3 hat i+10 hat j

Explanation:

The support line for force vec F_1 is given by

l_1->p = p_1+lambda_1 vec F_1

where p = {x,y}, p_1 = {1,0} and lambda_1 in RR.
Analogously for l_2 we have

l_2->p = p_2+lambda_2 vec F_2

where p_2 = {-3,14} and lambda_2 in RR.
The intersection point or l_1 nn l_2 is obtained equating

p_1+lambda_1 vec F_1 = p_2+lambda_2 vec F_2

and solving for lambda_1,lambda_2 giving

{lambda_1 = 2, lambda_2 = 2}

so l_1 nn l_2 is at {3,10} or 3 hat i+10 hat j

Jun 17, 2016

color(red)(3hati+10hatj)

Explanation:

Given

  • "The 1st force " vecF_1=hati +5hatj
  • "The 2nd force " vecF_2=3hati -2hatj
  • vecF_1 " acts at point A with position vector " hati
  • vecF_2 " acts at point B with position vector "-3 hati+14hatj

We are to find out the position vector of the point where the two given forces meet.

Let that point where the two given forces meet, be P with

position vector color (blue)(xhati+yhatj)

"Now displacement vector " vec(AP)= (x-1)hati+yhatj

"And displacement vector " vec(BP)= (x+3)hati+(y-14)hatj

"Since " vec (AP) and vecF_1 " are collinear we can write"

(x-1)/1=y/5=>5x-y=5......(1)

"Again " vec (BP) and vecF_2 " are collinear , so we can write"

(x+3)/3=(y-14)/-2=>2x+3y=36......(2)

Now multiplying equation (1) by 3 and adding with equation (2) we get

15x+2x=3xx5+36=>x=51/17=3

Inserting the value of x in equation (1)

5xx3-y=5=>y=10

"Hence the position vector of the point where the two given forces meet is " color(red)(3hati+10hatj)