Another hard M1 Edexcel Vector Question - Any takers??

Two forces F1 = (2i + 3xj) N and F2 = (xi + yj) N, where x and y are scalars, act on a particle.
The resultant of the two forces is R, where R is parallel to the vector i + 2j.

(a) Find, to the nearest degree, the acute angle between the line of action of R and the vector i.

(b) Show that 2x - y + 1 = 0.

Given that the direction of F2 is parallel to j,

(c) find, to 3 significant figures, the magnitude of R.

can any one to part b?

solutions are as follows:

(a) 63°

(c) 4.47 N

Reply 1

PrincessParadox

I tend to do these questons like gradients but i'm not getting the correct answer in this case, so someone please point out where im going wrong

3x+y =1
2+x =2

cross multiply to get 5x+2y-1=0

Reply 2

john !!

14

KingAS

Two forces F1 = (2i + 3xj) N and F2 = (xi + yj) N, where x and y are scalars, act on a particle.
The resultant of the two forces is R, where R is parallel to the vector i + 2j.

(a)Find, to the nearest degree, the acute angle between the line of action of R and the vector i.

(b)Show that 2x - y + 1 = 0.

Given that the direction of F2 is parallel to j,

(c) Find, to 3 significant figures, the magnitude of R.

(a) tanθ = 2/1
θ = 63.4

(b) Wtf...

Reply 3

nooneknows

Are you sure you've copied down the question correctly? It seems impossible to get x and y values which satisfy the requirements of R being parallel to i+2j, 2x-y+1=0 and xi+yj being vertical.

Correct me if I'm wrong, but for xi+yj to be vertical then x has to equal zero, if x=0 then 2x-y+1=0 means y=1 but then R is 2i+j, which certainly ain't parallel with i+2j!

If I'm doing something stupid, sorry! It just seems like an impossible/contradictory question.

Reply 4

Bezza

2

a) Since R is parallel to i + 2j, draw a triangle, tan(theta) = 2, theta = arctan(2) = 63°
b) I think you've made a mistake and F1 should be (2i + 3j) (no x). The resultant force is F1 + F2 = (2+x)i + (3+y)j = k(i+2j) So 2+x = k, 3+y = 2k, equating gives 2+x = (3+y)/2, 4+2x=3+y, so 2x - y + 1 = 0 (1)
c)F2 is parallel to j so i component zero ie x=0. R = 2i + (3+y)j, y = 1 from eqn (1) so R = 2i + 4j, |R| = sqrt(2^2 + 4^2) = sqrt(20) = 2sqrt5 = 4.47N

Reply 5

stanny

Bezza

a) Since R is parallel to i + 2j, draw a triangle, tan(theta) = 2, theta = arctan(2) = 63°
b) I think you've made a mistake and F1 should be (2i + 3j) (no x). The resultant force is F1 + F2 = (2+x)i + (3+y)j = k(i+2j) So 2+x = k, 3+y = 2k, equating gives 2+x = (3+y)/2, 4+2x=3+y, so 2x - y + 1 = 0 (1)
c)F2 is parallel to j so i component zero ie x=0. R = 2i + (3+y)j, y = 1 from eqn (1) so R = 2i + 4j, |R| = sqrt(2^2 + 4^2) = sqrt(20) = 2sqrt5 = 4.47N