Question

What can you say about the motion of the lift if the scales show a weight ofA)680NB)600N Describe the lifts motion in as much detail as you can

Answer 1

Given data:

* The weight of the woman is 680 N.

Solution:

The normal force acting on the woman shows the reading of the scale.

The net force acting on the woman showsthe motion of the lift.

The diagrammatic representation of the given case is,

The normal acting on the woman in terms of the net force and weight of the woman is,

`F_{\text{N}}-W=F_{\text{net}}`

Where W is the weight, F_net is the net force or accelerating force on the elevator, F_N is the normal force that shows the reading on the scale.

(a). When the reading on the scale is same as the weight of the women, then,

`F_N=W`

Thus, the net force acting on the lift is,

`\begin{gathered} W-W=F_{\text{net}} \\ F_{\text{net}}=0\text{ N} \end{gathered}`

According to the Newton's second law,

`F_{\text{net}}=ma`

where m is the mass, and a is the acceleration,

In this case,

`\begin{gathered} F_{\text{net}}=0 \\ ma=0 \\ a=0ms^(-2) \end{gathered}`

Thus, the lift is either moving with constant speed or the lift is in the state of rest.

(b). If the reading on the scale is,

`F_N=600\text{ N}`

Then, the net force acting on the lift is,

`\begin{gathered} F_N-W=F_{\text{net}} \\ 600-680=F_{\text{net}} \\ F_{\text{net}}=-80\text{ N} \end{gathered}`

The mass of the woman is,

`\begin{gathered} W=mg \\ m=(W)/(g) \end{gathered}`

where g is the acceleration due to gravity,

`\begin{gathered} m=(680)/(10) \\ m=68\text{ kg} \end{gathered}`

By Newton's second law, the net force acting on the lift in terms of the mass of the woman (considering the lift as the weightless )is,

`\begin{gathered} F_{\text{net}}=ma \\ -80=68* a \\ a=(-80)/(68) \\ a=-1.2ms^(-2) \end{gathered}`

If the lift is weightless than the acceleration with which the lift is moving is -1.2 meter per second squared.

Thus, the negative sign of the acceleration in this case represent either the lift is decelerating while moving in upward direction or lift is moving in the downward direction.