IIT JEE , AIEEE Forum - Physics , Mechanics

sneha_91

the sum of two forces is 8N.the magnitude of the resultant ,which is at right angles to the smaller force is 4N.find the force and the angle between them

sneha_91

please help me out

CHAMP007

Let the forces be X AND Y
X+Y=8
X^2+Y^2+2XYcos#=16
tan90=Ysin# / X +Ycos#
X+Ycos#=0
solving
we get Y=5 N
X=3N
cos# = -3/5

DO RATE ME IF CORRECT

kghedriu

well...let us assume the vectors to be F1 and 8-F1....now let us assume that F1 is the smaller vector......

for convenience i ll represent F1 by u and 8-F1 by v.....

now,

u^2+v^2+2uvcos

= 16 (formula for the resultant)

so

(u+v)^2+ 2uv+2uv cos

= 16

so, u+v=8 (given)...substituting...

48= 2uv(1-cos

)-------------------------(1)

now, angle made by the resultant is given as:

Tan

= v sin

/u+v cos

...and sice angle is 90 deg, it implies...

tan90 = infinity.....

therefore u+v cos

=0

.: cos

= -u/v-----------------------------(2)

substituting in equation (1)....we get....

48= 2uv(1-u/v)

now, put v=8-u....

u get 24= 8u...

.: u=3, hence V=5...

.: angle btween them = cos^-1(-3/5) (cos inverse of -3/5)

hope m clear..

thank u..

Goutham.K

ramyani

This shd be

u^2+v^2+2uvcos

= 16 (formula for the resultant)

so

(u+v)^2+ 2uv+2uv cos

= 16 + 2uv

so, u+v=8 (given)...substituting...

48= 2uv(1-cos

)-------------------------(1)
etc.

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