**Example 2.1** Find the components of the force 100 N along Sand Vas shown in Fig. P-2.1(n).

**Solution :**

F_{S} /sin 200 =100/sin 1100 F v sin 50^{0}

F_{S} = 100 sin 20^{0} / sin 1100 = 36 .397 (force along s ) Ans

F_{V} = sin 50^{0} /sin 100 x = 81.52 N( force along V ) Ans

**Example 2.2**. Fig. P-2.2(a) shows two forces. Fig. P-2.2(b) shows their resultant 3000 N acting at 15° as shown. Find out F_{1} and F_{2}.

**Fig. P-2.2(a) Fig. P-2.2(b)**

**Solution**. Resolving components along X and Y directions, we get

F_{1} cos 30° + F_{2} cos 45° = 3000 cos 15° ; and

– F_{1} sin 30° + F_{2 }sin 45° =-3000 sin 15°.

Solving the above two equations, we get

F_{1} = 2682 N. Ans.

F_{2} = 795 N. Ans.

**Example 2.3. **A force is having orthogonal components as 10 N, 20 N, and -30 N in x, y and z directions respectively. Determine (i) the magnitude of the force (ii) the direction cosines of the force.

**Solution:** Let the force be F and its components are F _{x}, F _{y} and F _{z}

F _{x}=10N,F _{y}=20N,F _{z}=-30N

Magnitude of

F= √( 10)^{2} + (20)^{2} + (- 30)^{2} = 37.416 N. Ans.

F_{ x} =f . l ; l =f _{x} /f =10 /37.416 N . Ans

F _{y} =F. m ; m =F _{y}/ F = 20/ 37.416 =0.535 Ans

F _{z} =F . n ; n =F _{z} /F = -30 /37.416 = – 0.802 Ans

**Example 2.4. **The 2000 N force is parallel to the displacement vector CB while the 1000 N force is parallel to the displacement vector OA. What is F_{1} + F_{2}.