4. Forces and Vectors

From Mechanics

F=Fcosi+Fcos(90−)j=Fcosi+Fsinj

Fcos is one component of the force. If i is horizontal, Fcos is called the horizontal component of the force.

Fsin is another component of the force. If j is vertical, Fsin is called the vertical component of the force.

Example 4.1

Express each of the forces given below in the form ai + bj.

Solution

(a)

20cos40i+20sin40j

(b)

−80cos30i+80sin30j

Note the negative sign here in the first term.

Example 4.2

Express the force shown below as a vector in terms of i and j.

Solution

28cos30i−28sin30j

Note the negative sign in the second term.

Example 4.3

Solution

−50cos44i−50sin44j

Note that here both terms are negative.

Example 4.4

Find the magnitude of the force (4i - 8j) N. Draw a diagram to show the direction of this force.

Solution

The magnitude, FN , of the force is given by,

F=42+82=80=894 N (to 3sf) 

The angle, , is given by,

=tan−148=634 

Example 4.5

Find the magnitude and direction of the resultant of the four forces shown in the diagram.

Solution

Force Vector Form 20 N 20cos50i+20sin50j

18 N −18j

25 N −25cos20i−25sin20j

15 N −15cos30i+15sin30j

\displaystyle \begin{align} & \text{Resultant Force }=\text{ }\left( 20\cos 50{}^\circ -25\cos 20{}^\circ -15\cos 30{}^\circ \right)\mathbf{i}+\left( 20\sin 50{}^\circ -18-25\sin 20{}^\circ +15\sin 30{}^\circ \right)\mathbf{j} \\ & =-23.627\mathbf{i}-3.730\mathbf{j} \end{align}

The magnitude is given by:

\displaystyle \sqrt{{{23.627}^{2}}+{{3.730}^{2}}}=23.9\text{ N (to 3sf)}

The angle \displaystyle \theta can be found using tan.

\displaystyle \begin{align} & \tan \theta =\frac{3.730}{23.627} \\ & \theta =9.0{}^\circ \end{align}