Solved problems in vectors – determine vector components

1. A force of 20 Newton makes an angle of 30^{o} with the x-axis. Find both the x and y component of the force.

Solution

F_{x} = F cos 30^{o} = (20)(cos 30^{o}) = (20)(0.5√3) = 10√3 Newton

F_{y} = F sin 30^{o} = (20)(sin 30^{o}) = (20)(0.5) = 10 Newton

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2. F_{1} = 20 Newton makes an angle of 30^{o} with the y axis and F_{2} = 30 Newton makes an angle of 60^{o} with the -x axis. Find both the x and y components of F_{1} and F_{2}.

Solution

F_{1x} = F_{1} cos 60^{o} = (20)(cos 60^{o}) = (20)(0.5) = -10 Newton (negative because it has same direction with -x axis)

F_{2x} = F_{2} cos 60^{o} = (30)(cos 60^{o}) = (30)(0.5) = -15 Newton (negative because it has same direction with -x axis)

F_{1y} = F_{1} sin 60^{o} = (20)(sin 60^{o}) = (20)(0.5√3) = 10√3 Newton (positive because it has same direction with y axis)

F_{2y} = F_{2} sin 60^{o} = (30)(sin 60^{o}) = (30)(0.5√3) = -15√3 Newton (negative because it has same direction with -y axis)

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3. F_{1} = 2 N, F_{2} = 4 N, F_{3} = 6 N. Find both the x and y components of F_{1}, F_{2} and F_{3}!

Solution

F_{1x} = F_{1} cos 60^{o} = (2)(cos 60^{o}) = (2)(0.5) = 1 Newton (positive because it has the same direction with x axis)

F_{2x} = F_{2} cos 30^{o} = (4)(cos 30^{o}) = (4)(0.5√3) = -2√3 Newton (negative because it has the same direction with -x axis)

F_{3x} = F_{3} cos 60^{o} = (6)(cos 60^{o}) = (6)(0.5) = 3 Newton (positive because it has the same direction with x axis)

F_{1y} = F_{1} sin 60^{o} = (2)(sin 60^{o}) = (2)(0.5√3) = √3 Newton (positive because it has the same direction with y axis)

F_{2y} = F_{2} sin 30^{o} = (4)(sin 30^{o}) = (4)(0.5) = 2 Newton (positive because it has the same direction with y axis)

F_{3y} = F_{3} sin 60^{o} = (6)(sin 60^{o}) = (6)(0.5√3) = -3√3 Newton (negative because it has the same direction with -y axis)

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