Question

Plz solve this physics problem

Answer 1

$\huge{\underline{ \boxed{ \bold{ \rm{ \purple{answer}}}}}}$

✏ Please see the attachment for answer...

By adding all equations we get...

$\implies \rm \: mgsin60 \degree + mgsin45 \degree + mgsin30 \degree = 4ma \\ \\ \therefore \rm \: g(sin60 \degree + sin45 \degree + sin30 \degree) = 4a \\ \\ \therefore \rm \: g( \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} } + \frac{1}{2} ) = 4a \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \pink{a} = \orange{ \frac{g}{4} \blue{( \red{ \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} }+ \frac{1}{2}})}} }}}} \: \purple{ \clubsuit}$

Answer 2

Answer:

Given:

A pulley system is given with 4 masses.

To find:

Acceleration of the system of masses.

Concept:

Draw the FBD of all the masses and try to find out their forces and acceleration. Finally solve the equations and get the acceleration value.

Calculation:

Going from bottom to upwards :

For 1st mass :

$mg \sin(60 \degree) - T1 = ma$

For 2nd mass :

$T1 + mg \sin(45 \degree) - T2 = ma$

For 3rd mass :

$T2 + mg \sin(30 \degree) - T3= ma$

For 4th mass :

$T3 = ma$

Adding all the equations, we see that all the Tension terms are cancelled and only acceleration remains.

Putting the value of sin function , and adding :

$= > mg \dfrac{ \sqrt{3} }{2} + mg \dfrac{1}{ \sqrt{2} } + mg \dfrac{1}{2} = 4ma$

Cancelling mass (m) on both sides :

$= > 4a = \bigg( \dfrac{ \sqrt{3} }{2} + \dfrac{1}{ \sqrt{2} } + \dfrac{1}{2} \bigg)g$

$= > a = \dfrac{g}{4} \bigg( \dfrac{ \sqrt{3} }{2} + \dfrac{1}{ \sqrt{2} } + \dfrac{1}{2} \bigg)$

So final answer :

$\boxed{ \red{ \huge{ \bold{ \sf{ \underline{Option \: \: B)}}}}}}$