InterviewSolution
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InterviewSolution
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A uniform metallic rod PQ of length 2 m is acted upon by two forces A and B along the directionsas shown in the figure. Find the magnitude and position of the resultant normal force that actson the rod. |
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Answer» Solution :The forces that act normally to the rod are ` A cos 60^(@) = 90 xx (1)/(2)=45N` ` B cos 60^(@) = 30 xx (1)/(2) = 15 N`. As the two forces are unlike parallel forces, the resultant of these two forces is outside the rod at a point 'O' and in the direction of greater force, i.e. , A.  Magnitude of R = 45 - 15 = 30 N POSITION of R: Let 'R' is at 'O' andata distance of X m from'P'. Then ` 45 xxx=15xx(2+x)` `3x=2+x` `2x=2` ` x = 1 m ` |
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