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Question 6
A uniform rod AB has length 3 m and weight 120 N. The rod rests in equilibrium in a horizontal position, smoothly supported at points C and D, where AC = 0.5 m and A... show full transcript
Step 1
Answer
To solve for W, we consider moments about point A:
The weight of the rod acts at its center, which is 1.5 m from point A. Therefore, the moment due to the weight of the rod is (120 \times 1.5).
The reaction at point C, which we'll denote as R, must balance the moments:
[ 3R + 120 \times 1.5 = R \times 2 + 2R \times 1 ] (where 2R captures the moment arm for the reaction at D).
This leads to:
[ 3R - R + 120 \times 1.5 = 0 ]
Rearranging gives:
[ 3R + 180 - 3R = W ]
Substituting W into the equation yields:
[ W(1 - x) = 60 ]
Finally, we arrive at:
[ W = \frac{60}{1 - x} ]
Step 2
Answer
From the equation (W = \frac{60}{1 - x}), we need to ensure that W is positive:
For (W > 0), we require (1 - x > 0), which simplifies to (x < 1).
Additionally, since W represents a weight, it must also be positive, implying (x > 0).
Therefore, combining these inequalities, we deduce the range for x to be:
[ 0 < x < 1 ]
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