Question

Determine the acceleration of the masses w.r.t. lift and tension in the string if the whole system in moving vertically upwards with uniform acceleration a0​. (m1​>m2​)

Answer 1

Answer:

let the tension be T

 m1(g+a0)−T=m1a 

T−m2(g+a0)=m2a

add both the equation , 

(m1−m2)(g+a0)=a(m1+m2) 

a=(m1+m2)(m1−m2)(g+a0)

 acceleration of the blocks wrt the lift.

 T+(m1+m2)m2(m1−m2)(g+a0)+m2(g+a0) tension in the string.

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Answer 2

Given:

• The lift has acceleration = ao

• The Masses are m1 and m2

• m1 > m2

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Need to Find:

• The acceleration of masses=?

• Tension in string =?

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Solution:

Let,

• The Tension be T
•  The acceleration of the blocks be a

Apply PseudoForce on the blocks 

The Pseudo Force:

=Mass of the body× Acceleration of the non-inertial frame

Now writing the equations of all the forces acting on the block m1 we get,

$\bold{\boxed{\purple{m_{1}g+m_{1}{a}_{0}-T={m}_{1}a-----i}}}$

Equation of all forces acting on block m2 is:

$\bold{\boxed{\purple{T-{m}_{2}(g+{a}_{o})={m}_{2}a-----ii}}}$

Adding the equation i and ii we get,

$({m}_{1}-{m}_{2})(g+{a}_{o})=a({m}_{1}+{m}_{2})$

$\red{\bold{\boxed{\large{a = \dfrac{({m}_{1} - {m}_{2})(g + {a}_{o})}{{m}_{1} + {m}_{2}}}}}}$

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Substitute the value of acceleration in equation 2, then we'll get,

$T= 2a+{m}_{2}{a}_{o}+{m}_{2}g$

$\implies T= {m}_{2} \times \dfrac{({m}_{1} - {m}_{2})(g + Ao)}{{m}_{1} + {m}_{2}} + {m}_{2}({a}_{o}+ g)$

$\red{\bold{\boxed{\large{T= \dfrac{2{m}_{1}{m}_{2}(g + {a}_{o})}{{m}_{1} + {m}_{2}}}}}}$