find the N-S and E-W components of the two forces
add the corresponding components to find the components of the resultant
The magnitude and direction of two forces acting on an object are 100 pounds, S79degreesE, and 60 pounds N68degreesE, respectively. Find the magnitude, to the nearest tenth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.
2 answers
The angles are referenced to the Y-axis. so we'll assume CW rotation from +Y-axis.
F1 = 100Lbs[79o] E. of S. = 180-79 = 101o CW,
F2 = 60Lbs[68o] E. of N. = 68o CW.
Fr = 100[101o] + 60[68o] = Resultant force,
X = 100*sin101 + 60*sin68 = 153.8 Lbs.
Y = 100*Cos101 + 60*Cos68 = 3.4 Lbs.
(Fr )^2 = X^2 + Y^2 = 23,666,
Fr = 153.84 Lbs. = Resultant force.
Tan A = X/Y For CW rotation.
F1 = 100Lbs[79o] E. of S. = 180-79 = 101o CW,
F2 = 60Lbs[68o] E. of N. = 68o CW.
Fr = 100[101o] + 60[68o] = Resultant force,
X = 100*sin101 + 60*sin68 = 153.8 Lbs.
Y = 100*Cos101 + 60*Cos68 = 3.4 Lbs.
(Fr )^2 = X^2 + Y^2 = 23,666,
Fr = 153.84 Lbs. = Resultant force.
Tan A = X/Y For CW rotation.
4 answers