An elevator cab of mass is suspended by a single vertical cable

PHYSICS · Multi-part · CBSE · Physics · Laws of Motion · Laws of Motion

Problem

An elevator cab of mass 1.2×103 kg1.2\times10^{3}\,\text{kg} is suspended by a single vertical cable. During three successive time intervals the upward force exerted by the cable on the cab has the following constant magnitudes: • Interval A (first 2.0 s2.0\,\text{s}): 1.18×104 N1.18\times10^{4}\,\text{N} • Interval B (next 3.0 s3.0\,\text{s}): 1.20×104 N1.20\times10^{4}\,\text{N} • Interval C (final 2.5 s2.5\,\text{s}): 1.05×104 N1.05\times10^{4}\,\text{N} Take the downward direction as positive and use g=9.8 m s−2g=9.8\,\text{m}\,\text{s}^{-2}. Assume the cable force changes instantaneously between intervals. (1) For each interval, calculate the net vertical force on the elevator cab, indicating its magnitude and direction. (2) For each interval, determine the elevator’s acceleration, including sign (direction). (3) Identify during which interval(s) the vertical forces on the cab are balanced. Justify your answer quantitatively. (4) Sketch a qualitative graph of the cab’s velocity versus time for the entire 7.5 s7.5\,\text{s}, starting from rest at t=0t=0. Label slopes and key features, and explain how the graph reflects the net-force conditions in each interval.

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