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?

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Answer 1 · 2016-06-16T11:06:08

#3 hat i+10 hat j#

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#

Answer 2 · 2016-06-17T12:13:24

#color(red)(3hati+10hatj)#

Given

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)#