Y = ver = -50 (given),
X^2 + (-50)^2 = (100)^2,
X^2 = 10000 - 2500 = 7500,
X = hor = sqrt(7500) = 86.6,
Vector to be added:
-50 + Y = 0,
Y = 50 added.
86.6 + X = -85,
X = hor = -85 - 86.6 = -171.6 added.
tanA = Y/X = 50/171.6 = 0.2914,
A = 16.24 deg.
Magnitude = y / sinA = 50/sin16.24
= 178.7
Direction = 16.24 deg above WEST.
You are given a vector in the xy plane that has a magnitude of 100.0 units and a y component of -50.0 units
(b) Assuming the x component is known to be positive, specify the vector which, if you add it to the original one would give a resultant vector that is 85.0 units long and points entirely in the -x direction.
Magnitude ?
Direction ?
2 answers
Correction:
Direction = 16.24 deg NORTH of WEST.
Direction = 16.24 deg NORTH of WEST.