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A uniform metre rod of 5 metre length is suspended horizontally by two strings P and Q.Pis attached 0.8 metre from one end and Q is attached 2 metre from the other end.Given that the weight of the rod is 110N. Calculate the tension in each string

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Answer:

The tension in string P is 25 N, while that of Q is 85 N.

Step-by-step explanation:

Considering the conditions for equilibrium,

i. Total upward force = Total downward force


T_(P) +
T_(Q) = 110 N

ii. Taking moment about P,

clockwise moment = anticlockwise moment

110 × (2.5 - 0.8) =
T_(Q) × (3 - 0.8)

110 × 1.7 =
T_(Q) × 2.2

187 = 2.2
T_(Q)


T_(Q) =
(187)/(2.2)


T_(Q) = 85 N

From the first condition,


T_(P) +
T_(Q) = 110 N


T_(P) + 85 N = 110 N


T_(P) = 110 - 85


T_(P) 25 N

Therefore, the tension in string P is 25 N while that of Q is 85 N.

User Michael De Soto
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