Question

A 5000kg elevator is to be designed so that the maximum acceleration is 10% og gravitational acceleration. What are the maximum and minimum forces the motor should exert on the supporting cable?

Answer 1

The maximum force is 55000 N,

The minimum force is 50000 N

Explanation:

Given as :

The mass of the elevator = m = 5000 kg

The gravitational acceleration = $a_g$ = 10 m/s²

The maximum acceleration = $a_m_a_x$ = 10% of $a_g$

Let The maximum force of motor = $F_m_a_x$  N

Let The minimum force of motor = $F_m_i_n$   N

According to question

∵ Force = Mass × acceleration

For maximum force

∵   $a_m_a_x$ = 10% of $a_g$

i.e  $a_m_a_x$ = 10% × 10 m/s²

So, $a_m_a_x$ = 1 m/s²

Now, Maximum force = mass ×  maximum acceleration

So, $F_m_a_x$ = 5000 kg × ($a_m_a_x$ + 10)

Or, $F_m_a_x$ = 5000 kg × 11 m/s²

∴     $F_m_a_x$ = 55000 N

Again

Minimum force = mass ×  minimum acceleration

So,  $F_m_i_n$= 5000 kg × $a_m_i_n$

Or,  $F_m_i_n$= 5000 kg × 10 m/s²  

∴     $F_m_i_n$ = 50000 N

Hence, The maximum force is 55000 N,

And The minimum force is 50000 N  Answer