Physics Paper 2, Nov/Dec. 2012
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
Weakness/Remedies
Strength

Question 2

Question 2
The upward acceleration of a lift of total mass 2500 kg is 0.5 ms-2.  The lift is
supported by a steel cable which has a maximum safe working stress of

1. x 107 Nm-2.  Determine the cross sectional area of the cable.

( g = 10ms-2 )

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comment

This question was not popular among the candidates and it was poorly attempted
Most candidates did not realize that for a body inside a upward moving lift it
experiences a weight increase hence they were unable to determine the tension
in the cable.  Some candidates stated that stress =   instead of stress =

Some candidates used  ‘g’ instead of (a+g).
Tension in the cable = (mass x acceleration) + (weight of the mass)
Stress, S  =            Tension
Cross-sectional area
OR
Cross-sectional
area                  =          mass (acceleration + acceleration of free fall)
stress
OR
A                     =          m (a+g)
A
=          2500 (0.5 + 10)
1.0 x 107

=          26.25 cm2        OR   2.625 x 10-3 m2