Question

Q1.A 1000 kg elevator is pulled up by a metallic wire that has maximum safe stress equal to 2 x 106 N/m2. If the diameter of the wire is 0.1 m, then the maximum safe acceleration of the elevator will be 1.9 m/s2 16 m/s2 5.7 m/s2 21 m/s2​

Answer 1

Given:

A 1000 kg elevator is pulled up by a metallic wire that has maximum safe stress equal to 2 x 10⁶ N/m2. The diameter of the wire is 0.1 m.

To find:

Max safe acceleration of the elevator?

Calculation:

First, let's calculate the total force (Tension) experienced by the wire while accelerating with a max acceleration of 'a'.

T = m(g + a)T=m(g+a)

\implies T = 1000(10 + a)⟹T=1000(10+a)

Now, area of wire = πd²/4 = π(0.1)²/4 = 0.0078 m².

So, max stress will be :

\therefore \: stress = \dfrac{T}{area}∴stress=

area

T

\implies \: 2 \times {10}^{6} = \dfrac{1000(10 + a)}{0.0078}⟹2×10

6

=

0.0078

1000(10+a)

\implies \: 2 \times {10}^{3} = \dfrac{(10 + a)}{0.0078}⟹2×10

3

=

0.0078

(10+a)

\implies \: 15.6 = 10 + a⟹15.6=10+a

\implies \: a =5.6 \: m {s}^{ - 2}⟹a=5.6ms

−2

\implies \: a \approx 5.7 \: m {s}^{ - 2}⟹a≈5.7ms

−2

So, max safe acceleration of the elevator is 5.7 m/s².

Answer 2

Step-by-step explanation:

Given:

A 1000 kg elevator is pulled up by a metallic wire that has maximum safe stress equal to 2 x 10⁶ N/m2. The diameter of the wire is 0.1 m.

To find:

Max safe acceleration of the elevator?

Calculation:

First, let's calculate the total force (Tension) experienced by the wire while accelerating with a max acceleration of 'a'.

T = m(g + a)

$\implies T = 1000(10 + a)$

Now, area of wire = πd²/4 = π(0.1)²/4 = 0.0078 m².

So, max stress will be :

$\therefore \: stress = \dfrac{T}{area}$

$\implies \: 2 \times {10}^{6} = \dfrac{1000(10 + a)}{0.0078}$

$\implies \: 2 \times {10}^{3} = \dfrac{(10 + a)}{0.0078}$

$\implies \: 15.6 = 10 + a$

$\implies \: a =5.6 \: m {s}^{ - 2}$

$</p><p>\implies \: a \approx 5.7 \: m {s}^{ - 2}$

So, max safe acceleration of the elevator is 5.7 m/s².