Determine the resultant of the system of forces:
Four forces act on a body as illustrated in Figure. Determine the resultant of the system of forces.
Figure
Solution
Resolving all of forces along x-axis, we attain
R_{x} = ∑ F_{x}
= F_{1} cos θ_{1} + F_{2} cos θ_{2} + F_{3} cos θ3 + F_{4 }cos θ_{4}
= 40 cos 30^{o} + 50 cos 315^{o} + 30 cos 180^{o} + 20 cos 240^{o}
Note: The angle created by 50 N forces is measured in anticlockwise direction from positive x axis after making the force work away from O by principle of transmissibility of the force.
∴ R_{x} = 40 cos 30^{o} + 50 cos 45^{o} - 30 cos 0^{o} - 20 cos 60^{o}
θ ≤ 90^{o} can be chosen in appropriate quadrant with correct signs as indicated above.
R_{x} = 34.64 + 35.36 - 30.00 -10.00
R_{x} = 30 N
Likewise, resolving all of the forces along y axis, we obtain
R_{y} = ∑ F_{y}
= F_{1} sin θ_{1} + F_{2} sin θ_{2} + F_{3} sin θ_{3} + F_{4} sin θ_{4}
= 40 sin 30^{o} + 50 sin 315^{o} + 30 sin 180^{o} + 20 sin 240^{o}
= 20.00 - 35.36 + 0.00 - 17.32
R_{y} = - 32.68 N
Therefore, the resultant in vector form can be expressed as
R¯ = (30 N) i¯ + (- 32.68 N) j¯
The magnitude of the resultant is specified by following
= 44.36 N
The direction θ may be worked out as,
θ = tan ^{-1} (R_{y} /R_{x})
= tan ^{-1} (- 32.68/30) = - 47^{o} 26′ 53′′
= 312^{o} 33′ 7′′
The resultant contain a magnitude of 44.36 N and is working in IVth quadrant making an angle of 312^{o} 33′ 7′′ in anticlockwise direction from positive x axis.