Class 11PhysicsNewton's Laws of Motion and FrictionMotion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)
MediumTwo masses of $5\, kg$ and $3\, kg$ are suspended with the help of massless inextensible strings as shown in the figure. Calculate $T_1$ and $T_2$ when the whole system is moving upwards with an acceleration $a = 2\, m/s^2$ (use $g = 9.8\, m/s^2$).
(N/A) Given: $m_1 = 5\, kg$,$m_2 = 3\, kg$,$g = 9.8\, m/s^2$,and $a = 2\, m/s^2$.
For the block of mass $m_2 = 3\, kg$:
The forces acting are tension $T_2$ upwards and weight $m_2g$ downwards. The net force is $T_2 - m_2g = m_2a$.
$T_2 = m_2(g + a) = 3(9.8 + 2) = 3(11.8) = 35.4\, N$.
For the block of mass $m_1 = 5\, kg$:
The forces acting are tension $T_1$ upwards,and tension $T_2$ and weight $m_1g$ downwards. The net force is $T_1 - T_2 - m_1g = m_1a$.
$T_1 = T_2 + m_1(g + a) = 35.4 + 5(9.8 + 2) = 35.4 + 5(11.8) = 35.4 + 59 = 94.4\, N$.
Thus,$T_1 = 94.4\, N$ and $T_2 = 35.4\, N$.
For the block of mass $m_2 = 3\, kg$:
The forces acting are tension $T_2$ upwards and weight $m_2g$ downwards. The net force is $T_2 - m_2g = m_2a$.
$T_2 = m_2(g + a) = 3(9.8 + 2) = 3(11.8) = 35.4\, N$.
For the block of mass $m_1 = 5\, kg$:
The forces acting are tension $T_1$ upwards,and tension $T_2$ and weight $m_1g$ downwards. The net force is $T_1 - T_2 - m_1g = m_1a$.
$T_1 = T_2 + m_1(g + a) = 35.4 + 5(9.8 + 2) = 35.4 + 5(11.8) = 35.4 + 59 = 94.4\, N$.
Thus,$T_1 = 94.4\, N$ and $T_2 = 35.4\, N$.
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