X = hor = 4 + (-5) = -1N.
Y = ver = 10 + (-6)
R = sqrt(x^2+y^2) = sqrt(1+16) = 4.12
tanA = Y / X = 4 / -1 = -4,
A = -76 deg, CW = 180-76 = 104 deg =
Direction.
describe the force board experiment to show how to obtain the resultant of two non-parallel coplanar force acting at a point.four forces of magnitude 10N, 5N, 4N, AND 6N acts on an oblect in the direction north. West, east and south respectively.find the magnitude and direction of their resultant
3 answers
Determine the magnitude and direction of the resultant of these force vectors having 20newton at20degrees north east,30newtons at 40degrees north west,40newtons at 45degrees south west and 10newtons at 70degrees south east
All angles are measured CCW from the +X-axis.
Fr = 20N[20o] + 30[140o[] + 40[225o] + 10[290o].
X = 20*Cos20 + 30*Cos140 + 40*Cos225 + 10*Cos290 = -29.05 N.
Y = 20*sin20 + 30*sin140 + 40*sin225 + 10*sin290 = -11.56 N.
Fr = sqrt(X^2 + Y^2 = 31.3 N.
TanA = Y/X.
A = 21.7o S. of W. = 201.7o CCW = Direction.
Fr = 20N[20o] + 30[140o[] + 40[225o] + 10[290o].
X = 20*Cos20 + 30*Cos140 + 40*Cos225 + 10*Cos290 = -29.05 N.
Y = 20*sin20 + 30*sin140 + 40*sin225 + 10*sin290 = -11.56 N.
Fr = sqrt(X^2 + Y^2 = 31.3 N.
TanA = Y/X.
A = 21.7o S. of W. = 201.7o CCW = Direction.
0 answers