Question

a steel cable by which an elevator of mass 800kg is suspended has a diameter of 2cm.the unstretched length of the suspension cable is 10m when the elevator is on the top floor of the building.the first floor is 50m below the top floor.find the values of extension in the cable when the elevator is on the top and on the first floor.

Answer 1

Answer on Question #59548, Physics / Mechanics | Relativity

a steel cable by which an elevator of mass $800\,\mathrm{kg}$ is suspended has a diameter of $2\,\mathrm{cm}$. The unstretched length of the suspension cable is $10\,\mathrm{m}$ when the elevator is on the top floor of the building. The first floor is $50\,\mathrm{m}$ below the top floor. Find the values of extension in the cable when the elevator is on the top and on the first floor.

Find: $\Delta l - ?\varepsilon - ?$

Given:

- $l = 60\,\mathrm{m}$

- $m = 800\,\mathrm{kg}$

- $d = 2 \times 10^{-2}\,\mathrm{m}$

- $E = 210 \times 10^{9}\,\mathrm{N/m^2}$

- $g = 9,8\,\mathrm{N/kg}$

Solution:

Hooke's Law: $\sigma = E\varepsilon (1)$,

where $\sigma$ – mechanical stress,

$E$ – Young's modulus,

$\varepsilon$ – relative elongation.

Mechanical stress: $\sigma = \frac{F_{\text{elast}}}{S} (2)$,

where $F_{\text{elast}}$ – elastic force,

$S$ – sectional area of the steel cable.

Elastic force numerically equal the weight of elevator:

$$F_{\text{elast}} = mg$$

Sectional area of the steel cable:

$$S = \frac{\pi d^2}{4}$$

(3) and (4) in (2):

$$\sigma = \frac{4\,mg}{\pi d^2}$$

Relative elongation: $\varepsilon = \frac{\Delta l}{l_0} (6)$,

where $\Delta l$ – absolute elongation,

$l_0$ – initial length of cable.

Absolute elongation:

$$\Delta l = l - l_0 ,$$

where $l$ – the final length of cable

(7) in (6): $\varepsilon = \frac{l - l_0}{l_0}$ (8)

(8) in (1): $\sigma = E \times \frac{l - l_0}{l_0}$ (9)

Of (5) and (9) $\Rightarrow \frac{4\mathrm{mg}}{\pi d^2} = E \times \frac{l - l_0}{l_0}$ (10)

Of (10) $\Rightarrow 4\mathrm{mgl}_0 = \mathrm{E}\pi \mathrm{d}^2 (l - l_0)$ (11)

Of (11) $\Rightarrow l_0(4\mathrm{mg} + \mathrm{E}\pi \mathrm{d}^2) = \mathrm{E}\pi \mathrm{d}^2 l$ (12)

Of (12) $\Rightarrow l_0 = \frac{E\pi d^2 l}{(4mg + E\pi d^2)}$ (13)

Of (13) $\Rightarrow l_0 = 59,9929\mathrm{m}$ (14)

(14) in (7): $\Delta l = 0,0071\mathrm{m}$ (15)

(14) and (15) in (6): $\varepsilon = 0,02\%$

**Answer:**

$\Delta = 7,1 \mathrm{~mm}$

$\varepsilon = 0,02\%$

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