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Asked by meenatoofansingh3 15th April 2021, 6:16 PM
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

..............................(4)

Hence , using eqn.(2) and (4) , we get tension T as

.......................(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 Expert 15th April 2021, 9:44 PM
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