answer

Asked by meenatoofansingh3 | 15th Apr, 2021, 06:16: PM

Expert Answer:

Qn.(6)
 
Let us consider 6N force is acting at D so that BD = x and AD ( 3 - x )
 
At equilibrium, sum of moments of forces is zero.
 
Let us take moments of forces about the point D.
 
At equilibrium, we have , 2 ( 3 - x ) = 4 x
 
Frome above expression, we get x = 1
 
Answer :- 6N force acting downward at a point D so that BD = 1 m
 
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Figure shows the tension force T acting along the string .
 
At the disk of mass M , tension force T is giving torque to rotate the disc with angular acceleration α
 
Hence we have,  T × R = I α .....................(1)
 
where R is radius of Disc , I = (1/2) M R2 is the moment of inertia and M is mass of disc
 
Angular acceleration α = a / R   ,
 
where a is linear accelertion at rim of disc that is same as acceleration of block of mass m
 
Hence we rewrite eqn.(1) as ,  T × R = (1/2) M R2 × ( a / R )
 
Hence we get , T = (1/2) M a   ...........................(2)
 
At the block of mass m , by applying Newton's second law , we get
 
( m g ) - T = m a  ...........................(3)
 
By adding eqn.(2) and (3) , we get  acceleration as
 
begin mathsize 14px style a space equals space fraction numerator m over denominator open parentheses m space plus space M over 2 close parentheses end fraction space g end style      ..............................(4)
 
Hence , using eqn.(2) and (4) , we get tension T as
 
begin mathsize 14px style T space equals space 1 half M cross times fraction numerator m over denominator open parentheses m space plus space begin display style M over 2 end style close parentheses end fraction space g end style     .......................(5)
 
Let us substitute values  M = 2.4 kg ,  m = 1.2 kg ,  g = 9.8 m/s2
 
We get tension T = (1/2) × 2.4 × [  1.2 / ( 1.2 + 1.2 ) ] × 9.8  = 5.88 N

Answered by Thiyagarajan K | 15th Apr, 2021, 09:44: PM

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