Find the reading of the spring balance shown in fig. The elevator is going up with an acceleration g/10, the pulley and the string are light and the pulley is smooth?( The mass of the blocks are 1.5 Kg and 3.5 Kg)

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2 Answers
Mar 3, 2018

4.4 Kg4.4Kg

Explanation:

Here,we can think from non inertial frame of reference that as the lift is moving upward with an acceleration of g/10g10,so net force acting downward for the two blocks will be their weight+(g/10)their mass.
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So,if the tension in the string is TT,we can write for the larger block as, 3g +(3g)/10 -T =3a3g+3g10T=3a (as it is going down)

similarly for the smaller block T-1.5g -(1.5g)/10 =1.5aT1.5g1.5g10=1.5a (where, aa is the acceleration of the blocks relative to the lift)

Solving both we get, T=(11/5)gT=(115)g

Now,the spring is attached to the pulley which is bearing the tension of the string from both the side,so the spring will show a reading of 2T=(22/5)g=43.12 N2T=(225)g=43.12N

so,reading of the spring balance will be ((22g)/5)/g =4.4 Kg22g5g=4.4Kg

Mar 3, 2018

See below.

Explanation:

Let the lift acceleration be gammaγ

x_1, m_1x1,m1 coordinate and mass of left body.
x_2,m_2x2,m2 coordinate and mass of right body.
x_0x0 the pulley center coordinate
LL rope length.

For left and right bodies

T-m_1 g = m_1 ddot x_1
T - m_2 g = m_2 ddot x_2

The restrictions to the movement are

(x_0-x_1) + (x_0-x_2) = L or

ddot x_0-ddot x_1+ddot x_0-ddot x_2 = 0 or

ddot x_0 = 1/2(ddot x_1 + ddot x_2)

Now solving

{(T-m_1 g = m_1 ddot x_1),(T - m_2 g = m_2 ddot x_2),(ddot x_0 = 1/2(ddot x_1 + ddot x_2)):}

we obtain

{(T = 2(m_1m_2)/(m_1+m_2)(ddot x_0 + g)),(ddot x_1 =(2 ddot x_0 m_2+(m_2-m_1)g)/(m_1+m_2) ),(ddot x_2 = (2 ddot x_0 m_1+(m_1-m_2)g)/(m_1+m_2)):}

but ddot x_0 = gamma then

T = 2(m_1m_2)/(m_1+m_2)(gamma + g)

So the spring balance will read

P = 2T = 4.4 g/g = 4.4