Question

Two blocks of mass 50 kg and 30kg Connected by a massless string pass over a light, friction-less pulley and rest on 2 smooth planes inclined at 30° and 60° respectively, with the horizontal. determined acceleration in the 2 blocks and tension in the string.​

Answer 1

Answer:

hope this ans helps you

Answer 2

Answer:

The tension and acceleration in the string are 576.47 N and 2.63 m/s²

Explanation:

Given that,

Mass of first block = 50 kg

Mass of second block = 30 kg

Inclined angle = 60°

Inclined angle for another box = 30°

In the both parts, the acceleration and tension are same in the string

$T(\sin30+\sin60)= (m_{1}+m_{2})g$...(I)

$T(\cos30-\cos60)=(m_{1}+m_{2})a$...(II)

We need to calculate the tension

Using balance equation

From equation (I)

Put the value into the formula

$T(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2})=(50+30)\times9.8$

$T=\dfrac{(50+30)\times9.8}{1.36}$

$T=576.47\ N$

We need to calculate the acceleration

Put the value of T in equation (II)

$576.47\times(\cos30-\cos60)=(50+30)a$

$a=\dfrac{211}{80}$

$a=2.63\ m/s^2$

Hence, The tension and acceleration in the string are 576.47 N and 2.63 m/s²