Answer
Verified
483.3k+ views
Hint: Make a free body diagram of the elevator and calculate the net force acting on the elevator. Calculate the maximum stress possible and hence the maximum force possible and apply the condition to the expression for the net force acting on the elevator to get max possible acceleration.
Complete step-by-step answer:
The given elastic limit of the elevator cable is $2\times 10^{9} N/m^2$
the area of cross-section of elevator cable is $10^{-4}m^2$
The mass of the elevator is $2\times 10^3$ kg
The free body diagram of the elevator can be drawn as
We are given the stress in the cable should not exceed half of the elastic limit of the cable so we get
We know the stress is given by $stress= \alpha = \dfrac{Tension}{Area} = \dfrac{T}{A} = \dfrac{m(g+a)}{a}$
We get from the given condition :
$\begin{align}
& \alpha =\dfrac{m(g+a)}{A}\le \dfrac{{{\alpha }_{max}}}{2}=\dfrac{2\times {{10}^{9}}}{2} \\
& g+a\le \dfrac{2\times 9}{2}\dfrac{{{10}^{-4}}}{2\times {{10}^{3}}}=50 \\
& a\le 50-10=40\dfrac{m}{{{s}^{2}}} \\
\end{align}$
Thus, using the free body diagram and calculating the net force acting on the elevator and the tension in the cable we got the maximum acceleration possible for the elevator as $40 \dfrac{m}{s^2}$.
Additional Information:
The net force acting on the elevator is equal to mass times the upward acceleration of the elevator as it is moving upward. Now we have seen that the net force in the upward direction is nothing but the T – mg where T is the tension in the rope. Hence we have equated this to the ma. So we get the equation T-mg = ma which gives us T = m(g+a) which we used in the problem.
Note: The possible mistake one can do in this kind of problem is that while taking the forces acting on the elevator. One should be able to differentiate between the forces acting on the body and the forces applied by the body on the cable.
Complete step-by-step answer:
The given elastic limit of the elevator cable is $2\times 10^{9} N/m^2$
the area of cross-section of elevator cable is $10^{-4}m^2$
The mass of the elevator is $2\times 10^3$ kg
The free body diagram of the elevator can be drawn as
We are given the stress in the cable should not exceed half of the elastic limit of the cable so we get
We know the stress is given by $stress= \alpha = \dfrac{Tension}{Area} = \dfrac{T}{A} = \dfrac{m(g+a)}{a}$
We get from the given condition :
$\begin{align}
& \alpha =\dfrac{m(g+a)}{A}\le \dfrac{{{\alpha }_{max}}}{2}=\dfrac{2\times {{10}^{9}}}{2} \\
& g+a\le \dfrac{2\times 9}{2}\dfrac{{{10}^{-4}}}{2\times {{10}^{3}}}=50 \\
& a\le 50-10=40\dfrac{m}{{{s}^{2}}} \\
\end{align}$
Thus, using the free body diagram and calculating the net force acting on the elevator and the tension in the cable we got the maximum acceleration possible for the elevator as $40 \dfrac{m}{s^2}$.
Additional Information:
The net force acting on the elevator is equal to mass times the upward acceleration of the elevator as it is moving upward. Now we have seen that the net force in the upward direction is nothing but the T – mg where T is the tension in the rope. Hence we have equated this to the ma. So we get the equation T-mg = ma which gives us T = m(g+a) which we used in the problem.
Note: The possible mistake one can do in this kind of problem is that while taking the forces acting on the elevator. One should be able to differentiate between the forces acting on the body and the forces applied by the body on the cable.
Recently Updated Pages
Master Class 9 Science: Engaging Questions & Answers for Success
Master Class 9 English: Engaging Questions & Answers for Success
Class 9 Question and Answer - Your Ultimate Solutions Guide
Master Class 9 Maths: Engaging Questions & Answers for Success
Master Class 9 General Knowledge: Engaging Questions & Answers for Success
Class 10 Question and Answer - Your Ultimate Solutions Guide
Trending doubts
Pigmented layer in the eye is called as a Cornea b class 11 biology CBSE
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE
Define cubit handspan armlength and footspan class 11 physics CBSE
Maximum speed of a particle in simple harmonic motion class 11 physics CBSE
Give a brief account on the canal system in sponge class 11 biology CBSE
Assertion Pila has dual mode of respiration Reason class 11 biology CBSE