Question

b) A uniform metal tube of length 5m and mass 9kg is suspended
honizontally two vertical wires atteched at 50m and 150m
respectively from the ends of the tube. Find the tension in each wire​

Answer 1

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Answer 2

Answer:

Given: 5 m (500 cm) tube with a mass of 9 kg suspended vertically from 150 cm and 50 cm from each end.

Total force F = ma, ==> F = 9.8 x 9 = 88.2 N

Note; this bar is static; upward forces = downward forces

From center(250 cm) to the left 100 cm a vertical wire and 200 cm to the right of center another vertical wire.

To be balance Force(F) x Distance(D) on both side of center must be equal.

F x D = F x D

F x 100 = F x 200

F = 2F

These forces will equal total force

F + 2F = 88.2

F(1 + 2) = 88.2

3F = 88.2

F = 29.4 and 2F = 58.8 N

So, Vertical wire 100 cm left of center will have a force (tension) of 29.4 N (upward)

Vertical wire 200 cm right of center will have a force (tension) of 58.8 N (upward).