Example 2.1 Find the components of the force 100 N along Sand Vas shown in Fig. P-2.1(n).
FS /sin 200 =100/sin 1100 F v sin 500
FS = 100 sin 200 / sin 1100 = 36 .397 (force along s ) Ans
FV = sin 500 /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 F1 and F2.
Fig. P-2.2(a) Fig. P-2.2(b)
Solution. Resolving components along X and Y directions, we get
F1 cos 30° + F2 cos 45° = 3000 cos 15° ; and
- F1 sin 30° + F2 sin 45° =-3000 sin 15°.
Solving the above two equations, we get
F1 = 2682 N. Ans.
F2 = 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
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 F1 + F2.