Solve the following :

Question:

Find the reading of the spring balance shown in figure. The elevator is going up with an acceleration of $g / 10$, the pulley and the string are light and the pulley is smooth.

Solution:

$\mathrm{T}-1.5 \mathrm{~g}-\frac{\frac{1.5 \mathrm{~g}}{10}}{10}=1.5 \mathrm{a} \mathrm{T}-3 \mathrm{~g}-\frac{3 \mathrm{~g}}{10}=-3 \mathrm{a}$

$\mathrm{T}=1.5 \mathrm{~g}^{\frac{1.5 \mathrm{~g}}{10}}+1.5 \mathrm{a} \mathrm{T}=3 \mathrm{~g}^{\frac{3 \mathrm{~g}}{10}}-3 \mathrm{a}$

$1.5 g^{\frac{1.5 g}{10}}+1.5 a=3 g_{-10}^{\frac{3 g}{10}}-3 a$

$4.5 \mathrm{a}=\frac{11 \mathrm{~g}}{10}$

Force on spring

Balance $=2 \mathrm{~T}$

$=2 \times\left(1.5 a+\frac{1.5 \times 11 g}{10}\right)$

$=2 \times\left(1.5 \times 3.593+\frac{1.5 \times 11 g}{10}\right)$

$=43.1 \mathrm{~N}$

Mass (reading ) of spring $=\frac{F}{g}$

$=\frac{2 T}{g}$

$43.1$

$=9.8=4.4 \mathrm{~kg}$

 

Leave a comment

None
Free Study Material