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In equilbrium,

net force = 0

Let T be the tension of the string,

In y-direction,

CodeCogsEqn (8)

..................................(1)

In x-direction,

CodeCogsEqn (7)

....................................(2)

By solving (1) and (2), we can get F = 8.5 (corr to 1 d.p.)

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