A uniform beam, PQ. is 100 m long and weighs 35 N. It is placed on a support at a point 40 cm from P. If weights of 54 N and FN are attached at P and Q respectively in order to keep it in a horizontal position, calculate, correct to the nearest whole number, the value of F.

54 x 40 = 35 x 10 + 60F
160 = 350 + 60F
F = \(\frac{1810}{60}\)
= 30N
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