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A black weighing W = 10 kN is resting on an inclined plane as shown in Determine its components normal to and parallel to the inclined plane. |
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Answer» 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. 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 Component parallel to the plane = W sin 20° = 10 sin 20° = 3.42 kN, down 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. |
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