Example 1.3. A black weighing W = 10 kN is resting on an inclined plane as shown in Fig. 1.10(a). Determine its components normal to and parallel to the inclined plane.
A
70
W
20°
(a)
B
20°
C
(b)
Fig. 1.10
Solution: The plane makes an angle of 20° to the horizontal. Hence the normal to the plane makes an angles of 70° to the horizontal i.e., 20° to the vertical [Ref. Fig. 1.10(b)]. If AB represents the given force W to some scale, AC represents its component normal to the plane and CB represents its component parallel to the plane. Thus from ∆ABC, Component normal to the plane = AC W cos 20° = 10 cos 20° = 9.4 kN as shown in Fig. 1.10(b) W sin 20° 10 sin 20° = 3.42 kN, down the plane Component parallel to the plane From the above example, the following points may be noted:
1. Imagine that the arrow drawn represents the given force to some scale.
2. Travel from the tail to head of arrow in the direction of the coordinates selected.
3. Then the direction of travel gives the direction of the component of vector.
4. From the triangle of vector, the magnitudes of components can be calculated.
Example 1.4. The resultant of two forces, one of which is double the other is 260 N. If the direction of the larger force is reversed and the other remain unaltered, the magnitude of the resultant reduces to 180 N. Determine the magnitude of the forces and the angle between the forces.
Solution: Let the magnitude of the smaller force be F. Hence the magnitude of the larger force is 2F. Thus F₁F and F₂ = 2F Let 9 be the angle between the two forces. From the condition 1, we get RF²+2FF₂ cos 0 + F22 = 260 F² + 2F (2F) cos + (2F)2 = 2602 5F2 + 4F2 cos 0 = 67600 i.e.. ...(i)
