A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is hanging from the board. The tension in the right chain is 250 N. What is the tension in the left chain?
Correct answer: C. 375 N
- A. 125 N
- B. 250 N
- C. 375 N
- D. 625 N
Explanation
To find the tension in the left chain, consider the principles of equilibrium. The total weight supported by the chains is the sum of the board's and the person's weights (125 N + 500 N = 625 N). Given that the tension in the right chain is 250 N, the left chain must support the remaining weight: 625 N - 250 N = 375 N. The distribution of weight is not equal because the person is hanging, creating a rotational effect, which requires balancing the torques for equilibrium.
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