Advertisement Remove all ads

Find the Reading of the Spring Balance Shown in Figure (5−E6). the Elevator is Going up with an Acceleration G/10, the Pulley and the String Are Light and the Pulley is Smooth. - Physics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

Find the reading of the spring balance shown in the following figure. The elevator is going up with an acceleration g/10, the pulley and the string are light and the pulley is smooth.

Advertisement Remove all ads

Solution

Let the left and right blocks be A and B, respectively.
And let the acceleration of the 3 kg mass relative to the elevator be 'a' in the downward direction.


From the free-body diagram,
\[m_A a = T - m_A g - \frac{m_A g}{10} . . . \left( 1 \right)\]
\[m_B a = m_B g + \frac{m_B g}{10} - T . . . \left( 2 \right)\]
Adding both the equations, we get:
\[a\left( m_A + m_B \right) = \left( m_B - m_A \right)g + \left( m_B - m_A \right)\frac{g}{10}\]
Putting value of the masses,we get:
\[9a = \frac{33g}{10}\]
\[ \Rightarrow \frac{a}{g} = \frac{11}{30} . . . \left( 3 \right)\]
Now, using equation (1), we get:
\[T = m_A \left( a + g + \frac{g}{10} \right)\]
The reading of the spring balance =\[\frac{2T}{g} = \frac{2}{g} m_A \left( a + g + \frac{g}{10} \right)\]
\[\Rightarrow 2 \times 1 . 5\left( \frac{a}{g} + 1 + \frac{1}{10} \right) = 3\left( \frac{11}{30} + 1 + \frac{1}{10} \right)\]
= 4 . 4 kg

Concept: Newton’s Second Law of Motion
  Is there an error in this question or solution?

APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 5 Newton's Laws of Motion
Q 16 | Page 80
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×