Physics, asked by AstroAlpha178, 9 months ago

Please answer this Its a question related to pulleys, NLM and constraints i guess

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Answers

Answered by dkdevender
0

Answer:

sorry I didn't know this answer

Answered by Anonymous
0

Method 1: (Ground frame):

Let acceleration of block m  

1

​  

 with respect to pulley to a upward and the acceleration of m  

2

​  

 w.r.t. pulley is a (downward)  

Equation of motion for  

m  

1

​  

 

           T−m  

1

​  

g=m  

1

​  

a  

1

​  

..........(i)

          T−m  

2

​  

g=m  

2

​  

a  

2

​  

..........(ii)

           

a

 

1

​  

=  

a

 

1.p

​  

+  

a

 

p

​  

=a+a  

0

​  

..........(iii)

           

a

 

2

​  

=  

a

 

2.p

​  

+  

a

 

p

​  

=−a+a  

0

​  

.........(iv)

Substituting a  

1

​  

 from (iii) in (i),

           T−m  

1

​  

g=m  

1

​  

(a+a  

0

​  

)........(v)

Substituting a  

2

​  

 from (iv) in (ii)

          T−m  

2

​  

g=m  

2

​  

(−a+a  

0

​  

).......(vi)

Solving (v) and (iv),

         T=  

m  

1

​  

+m  

2

​  

 

2m  

1

​  

m  

2

​  

 

​  

(g+  

​  

)

and   a=  

m  

1

​  

+m  

2

​  

 

m  

2

​  

−m  

1

​  

 

​  

(g+a  

0

​  

)

Method 2 : Solving problem from non-inertial frame of reference

Let us build the equations by using Newton's second  law sitting on the accelerating pulley. Hence, we impose pseudo force m  

1

​  

a  

0

​  

↓ and m  

2

​  

a  

0

​  

↓ on both m  

1

​  

 and m  

2

​  

, respectively, in addition to the upward tension and their weights m  

1

​  

g↓m  

2

​  

g↓, respectively, If m  

1

​  

 accelerates up relative to the pulley, m  

2

​  

 must accelerate down relative to the pulley with acceleration a.

Force equation :

For m  

1

​  

:T−m  

1

​  

g−m  

1

​  

a  

0

​  

=m  

1

​  

a........(i)

For m  

2

​  

:Tm  

2

​  

g+m  

2

​  

a  

0

​  

−T=m  

2

​  

a........(ii)

Solving (i) and(ii), we have  

        a=  

m  

1

​  

+m  

2

​  

 

m  

2

​  

−m  

1

​  

 

​  

(g+a  

0

​  

)

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