The forces 𝐹 one, which equals negative four 𝑖 plus six 𝑗 newtons; 𝐹 two, which is equal to negative nine 𝑖 plus four 𝑗 newtons; and 𝐹 three, which is equal to negative four 𝑖 minus three 𝑗 newtons are acting on a particle, where 𝑖 and 𝑗 are two perpendicular unit vectors. Determine the magnitude of the forces resultant, 𝑅, and its direction to the nearest minute.
Well, the resultant force is equal to 𝐹 one plus 𝐹 two plus 𝐹 three. This is equal to negative four 𝑖 plus six 𝑗 minus nine 𝑖 plus four 𝑗 minus four 𝑖 minus three 𝑗. Grouping the 𝑖 terms gives us negative 17𝑖. And grouping the 𝑗 terms gives us plus seven 𝑗. Therefore, the resultant force is negative 17𝑖 plus seven 𝑗.
The magnitude of the resultant is calculated by square-rooting negative 17 squared plus seven squared. We square the coefficient of 𝑖 and the coefficient of 𝑗, add the two answers, and then square-root this number. Negative 17 squared plus seven squared is 338. And the square root of 338 is 13 root two. This means that the magnitude of the resultant of forces 𝐹 one, 𝐹 two, and 𝐹 three is 13 root two newtons.
The direction of this resultant or angle 𝜃 is equal to the inverse tan of seven divided by negative 17. This is the 𝑗 coefficient divided by the 𝑖 coefficient. This is equal to negative 22.38. However, as we want a positive angle, we can add 180 degrees to this. This gives us 157.62.
Our final step is turning this into degrees and minutes. Therefore, 𝜃 equals 157 degrees and 37 minutes, as 0.62 multiplied by 60 is 37 minutes. We can therefore say that the direction of the resultant is 157 degrees 37 minutes. This can also be demonstrated on the diagram where 𝑅 is the resultant force and 𝜃 is the angle or direction.
Using Pythagoras’s theorem and trigonometry, we can calculate once again that the magnitude of the force is 13 root two newtons. And the direction 𝜃 is 157 degrees 37 minutes.