Question

In an ideal Atwood machine having only two blocks on either side of pulley, the sum of two masses is constant. If the string can sustain a tension equal to 24 50 of the weight of the sum of the two masses, then the least acceleration of the masses would be (in ms -2) (g=10ms-2)​

Answer 1

50−T1 = 5a

T1 −30 = 3a

T1  = 37.5

it is not depends on horizontal acceleration.

Answer 2

Given:

An ideal Atwood machine with two blocks and a pulley

The sum of the two masses is constant

The maximum tension that can be sustained by the string =  24/50 (sum of masses)

To Find:

The least acceleration (a) of the masses

Solution:

Let the two masses be m and M and the tension in the string be T.

Since the Atwood machine is an ideal one, we can write the force equations as follows:

For mass m: mg - T = ma       -(1)

For mass M: Mg- T = Ma         -(2)

Solving the above equations using the substitution method,

a = $\frac{m - M}{m + M} .g$         - (3)

and T = $\frac{2mM}{m + M} .g$     -(4)

According to the question,

T ≤ 24/50 (m + M)

or $\frac{2mM}{m+M} .g \leq \frac{12}{25} (m + M)$        -(5)

Cross multiplying and solving,

50 X m X M X g ≤ 12 X (m+M)²

Since the masses are constant,

⇒ the acceleration of the system is minimum when T is maximum

So, 50 X m X M X g = 12 X (m+M)²

Taking g = 10m/s² and solving,

500mM = 12 (m+M)²

or 125 [ (m+M)² - (m - M)² ] = 12 (m+M)²

or 125 (m+M)²  - 125 (m - M)² = 12 (m+M)²

or 113 (m+M)² = 125 (m - M)²

or (m+M)² /  (m - M)²  = 125 / 113

Taking the square root of both the sides,

m + M / m - M = $\sqrt{1.1}$

From equation (3),

m + M / m - M = g / a

or g / a = $\sqrt{1.1}$

or a = 10 / $\sqrt{1.1}$ ≈ 10 m/s²

Hence, the least value of acceleration for the given system is 10 m/s².