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Question: In the figure shown neglecting friction and mass of pulley what is the acceleration of mass \(B\). ...

In the figure shown neglecting friction and mass of pulley what is the acceleration of mass BB.

A. g3\dfrac{g}{3}
B. 5g2\dfrac{{5g}}{2}
C. gg
D. 2g5\dfrac{{2g}}{5}

Explanation

Solution

In this problem we need to find the acceleration of the block BB which is having the mass mm in terms of acceleration due to gravity by considering the friction and mass of the pulley as zero. The tension needs to be marked in the given pulley systems. We know that the tensions are the pulling force whose direction is always away from the load as shown in the figure below. The given pulley system consists of one fixed pulley and one movable pulley.

Complete step by step answer:

2aA=aB2{a_A} = {a_B} ………. (1)\left( 1 \right)
From block AA with the mass mm
mgTA=maAmg - {T_A} = m{a_A}
On simplifying the above equation we get,
mg=maA+TAmg = m{a_A} + {T_A} ……….. (2)\left( 2 \right)
From block BB with the mass mm
TBmg=maB{T_B} - mg = m{a_B} ……… (3)\left( 3 \right)
Substituting equation (1)\left( 1 \right) in equation (3)\left( 3 \right) we get
TBmg=m(2aB){T_B} - mg = m\left( {2{a_B}} \right)
On simplifying the above equation
TBmg=2maB{T_B} - mg = 2m{a_B}
Multiply both sides by22, we can write the above equation as
2TB2mg=4maB2{T_B} - 2mg = 4m{a_B}

On further simplification
2mg=2TB4maB2mg = 2{T_B} - 4m{a_B}……….. (4)\left( 4 \right)
Given that mass of the pulley is neglected
We have, TA=2TB{T_A} = 2{T_B} ………..(5)\left( 5 \right)
On subtracting equation (2)\left( 2 \right) and equation (5)\left( 5 \right) we get,
2mgmg=2TB4maAmaATA2mg - mg = 2{T_B} - 4m{a_A} - m{a_A} - {T_A} ……..(6)\left( 6 \right)
From equation (5)\left( 5 \right), TA=2TB{T_A} = 2{T_B} , equation (6)\left( 6 \right) becomes
mg=2TB4maAmaA2TBmg = 2{T_B} - 4m{a_A} - m{a_A} - 2{T_B}
Therefore on further simplification, we can write above equation as
mg=5maAmg = - 5m{a_A}
aA=g5\Rightarrow {a_A} = \dfrac{g}{5} ………..(7)\left( 7 \right)
We know that, aB=2aA{a_B} = 2{a_A}
Substituting equation (7)\left( 7 \right) in above equation, we get
aB=2g5\therefore {a_B} = \dfrac{{2g}}{5}

Hence, option D is correct.

Note: Single fixed pulley and single movable pulley are important pulleys which are used in mechanics. The ideal single fixed pulley has mechanical advantage of 11 whereas ideal movable pulley has mechanical advantage of 2. The fixed pulley was used to change the direction of the effort only.