# A 4 kg object is suspended by a string from the ceiling of an elevator. The acceleration of...

## Question:

A 4 kg object is suspended by a string from the ceiling of an elevator. The acceleration of gravity is 9.8 m/s{eq}^2 {/eq}.

Part 1

Determine the tension in the string if it is accelerating upward at a rate of 2.4 m/s{eq}^2 {/eq}. Answer in units of N. Part 2

Determine the tension in the string if it is accelerating downward at a rate of 2.4 m/s{eq}^2 {/eq}. Answer in units of N.

## Tension Inside an Accelerating Elevator

The three laws of Newton work fine in an inertial frame of reference. That is, in a frame moving with a constant velocity. However when mechanics is being done in an accelerated frame then unless fictitious forces (also called pseudo or inertial forces) are incorporated into the framework then Newton's laws are seen to break down. For eg, imagine that you are inside a train in uniform motion. Then a box placed on the floor remains where it is unless someone pushes or pulls it, thereby confirming Newton's first law. But now if the train accelerates forward, then an observer in the train will note that the box has started sliding backwards apparently of its own accord thereby violating the first law. The remedy is to introduce a fictitious force. In a frame accelerating at {eq}\displaystyle {a} {/eq} assign a mass {eq}\displaystyle {m} {/eq} a fictitious force {eq}\displaystyle {-ma} {/eq}. Now take into account all the real contact and long-range forces and apply Newton's laws to get the correct dynamics.

Here it is given that a 4 kg object is suspended by a string from the ceiling of an elevator. The acceleration due to gravity is {eq}\displaystyle {9.8\ m/s^2} {/eq}.

Part 1.

The elevator is accelerating upwards at {eq}\displaystyle {2.4\ m/s^2} {/eq}. So the forces acting on the object are the gravitational force {eq}\displaystyle {mg=4\times 9.8=39.2\ N} {/eq} vertically downwards, the fictitious force of magnitude {eq}\displaystyle {ma=4\times 2.4=9.6\ N} {/eq} again vertically downwards and finally the tension of the string holding out against the downward pull of these two forces. So at equilibrium,

{eq}\displaystyle {T=39.2+9.6=48.8\ N} {/eq}.

Part 2.

The elevator is now accelerating downwards at {eq}\displaystyle {a=2.4\ m/s^2} {/eq}.Therefore the fictitious force now acts vertically up. Hence the force balance condition becomes,

{eq}\displaystyle { T=39.2-9.6=29.6\ N} {/eq}. 