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Question 3
A uniform rod AB has length 1.5 m and mass 8 kg. A particle of mass m kg is attached to the rod at B. The rod is supported at the point C, where AC = 0.9 m, and the ... show full transcript
Step 1
Answer
To prove that , we start by analyzing the torque about point C. The rod's weight acts at its center of mass, which is at 0.75 m from A. Thus, the torque due to the rod is:
For the particle at B, the torque is given by:
Setting the counterclockwise torque equal to the clockwise torque for equilibrium, we have:
Simplifying, we find:
Solving for m gives:
m = rac{1.2}{0.6} = 2.
Step 2
Answer
To find the distance AD, we need to set up the moments around point D.
Let the distance between point A and D be . Then, the distance from D to B is .
The moment about point D will be:
Substituting the known values:
After canceling out , we get:
Expanding and rearranging gives:
This simplifies to:
Thus,
4.5x = 6 \\ x = rac{6}{4.5} = rac{4}{3} = 0.6 ext{ m}.
Therefore, the distance AD is:
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