Geomechanics and Ground Control · Engineering Mechanics · 2 marks · Numerical answer
A 600 N worker lifts 100 N as drawn. Half the worker’s weight acts at A. The lumbar joint L is a smooth hinge; erector-spinae force FE is parallel to x. Use the shown 25 cm and 40 cm horizontal moment arms and 5 cm muscle moment arm. Find resultant joint reaction in N, to two decimals.
Moment equilibrium about L gives FE×0.05=300×0.25+100×0.40, so FE=2300 N. The muscle direction is 35° from vertical as drawn. Its horizontal component is 2300sin 35°, while its downward component adds to the 400 N external vertical load. Hence reaction magnitude is√[(2300sin 35°)²+(2300cos 35°+400)²]≈2637.7 N.
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