Asked by Godson

Two forces 10 N acting in the direction N 30 east and 15 N acting in the eastern direction, if both forces act at a point, find the magnitude and direction of (a) the resultant force (b) the equilibriant force.
Answered by Henry
a. Fr = 10N[60o]CCW + 15N[0o]

X = 10*Cos60 + 15 = 20 N.
Y = 10*sin60 = 8.7 N.
Q1.

Tan A = Y/X = 8.7/20 = 0.433
A = 23.4o N. of E.

Fr = X/Cos A = 20/Cos23.4=21.8 N.[23.4o]
N. of E. = 23.4o CCW

b. Equilibrant:
X = -20
Y = -8.7
Q3.

Tan A = Y/X = -8.7/-20 = 0.433
A = 23.4o S. of W. = 203.4o CCW.

Fr = X/Cos A = -20/Cos203.4 = 21.8 N[23.4o S. of W. = 203.4o CCW.

Note: 23.4o S. of W. is the opposite of
23.4o N. of E. and their magnitudes are
equal.


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