(2) and (4) cancel each other
(1) and (3) result in 19 N @ 270º
find the net resultant
find the cancelling (balancing) force
... equal magnitude , opposite direction
Five forces act on an object.
(1) 56 N at 90° <-
(2) 40 N at 0°
(3) 75 N at 270° <-
(4) 40 N at 180°
(5) 50 N at 60°
What are the magnitude and direction of a sixth force that would produce equilibrium?
Answer to Find:
? N at ?°
2 answers
Fr = 56i + 40 - 75i -40 + (50*Cos60 + 50*sin60).
X = 40 - 40 + 50*Cos60 = 25N.
Y = 56 - 75 + 50*sin60 = -18.13 N.
Fr = sqrt(X^2 + Y^2) = 30.9 N.
TanA = Y/X, A = -35.95o = 35.95o S. of E. = 324o CCW.
The equilibrant force(Fe) must be equal in magnitude and 180o out phase with Fr.
Fe = 30.9N[35.95o] N. of W. = 30.9 N.[144.1o ] CCW.
X = 40 - 40 + 50*Cos60 = 25N.
Y = 56 - 75 + 50*sin60 = -18.13 N.
Fr = sqrt(X^2 + Y^2) = 30.9 N.
TanA = Y/X, A = -35.95o = 35.95o S. of E. = 324o CCW.
The equilibrant force(Fe) must be equal in magnitude and 180o out phase with Fr.
Fe = 30.9N[35.95o] N. of W. = 30.9 N.[144.1o ] CCW.