Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
a certain force has magnitude of 200N, its x component is -80N. what is its y component and its direction? two possible answers exist ... reply quickly..
F = 200 Fcos@ = Fx (x compponent )) ...............1 Fsin@ = Fy ( y component ) .............2 from 1 200cos@ = -80 cos@ = -2/5 here @ can be in 2nd or 3rd quadrant , if @ is in 2nd quadrant then cos@ is -ve, sin@ is +ve@ cos@ = -2/5 , sin@ = (1-cos2@)1/2 = (21)1/2/5 now , Fy = F sin@ = 200(21)1/2/5 = 184N if @ is in 3rd quadrant then cos@ is -ve , sin@ is -ve cos@ = -2/5 , sin@ = -(21)1/2/5 Fy = Fsin@ = -184N therefore two possible answers are 184N & -184N aprove if u like my ans
F = 200
Fcos@ = Fx (x compponent )) ...............1
Fsin@ = Fy ( y component ) .............2
from 1
200cos@ = -80
cos@ = -2/5
here @ can be in 2nd or 3rd quadrant ,
if @ is in 2nd quadrant then cos@ is -ve, sin@ is +ve@
cos@ = -2/5 , sin@ = (1-cos2@)1/2 = (21)1/2/5
now , Fy = F sin@ = 200(21)1/2/5 = 184N
if @ is in 3rd quadrant then cos@ is -ve , sin@ is -ve
cos@ = -2/5 , sin@ = -(21)1/2/5
Fy = Fsin@ = -184N
therefore two possible answers are 184N & -184N
aprove if u like my ans
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -