Practice examples of Newton’s laws and Atwood machine problems, calculation of acceleration using constraint.
Consider the given arrangement, as shown in figure.
In this arrangement, both blocks will have same acceleration because by visual inspection it can be seen that both blocks are connected with same string.
Acceleration of block B = Acceleration of block A
\(m_1g-T=m_1a\) ....(i)
(Since, no external force is applied, motion will take place only due to \(m_1g\) downward)
\(T=m_2a\) ...(ii)
(As only one force \(T\) is acting on block \(B\) in the direction of acceleration)
\(m_1g-m_2a=m_1a\)
\(m_2a+m_1a=m_1g\)
\(a=\dfrac{m_1g}{m_2+m_1}\)
Tension in the string
\(T=\dfrac{m_2.m_1g}{m_1+m_2}\)
A \(10\,m/s^2\)
B \(6.66\,m/s^2\)
C \(12\,m/s^2\)
D \(8\,m/s^2\)
Free body diagrams
For block A
\(T-m_1g=m_1a\) ... (i)
For block B
\(m_2g-T=m_2a\) ... (ii)
\(T=(m_1a+m_1g)\) {By solving equation (i)}
Putting value of T in equation (ii)
\(m_2g-m_1g-m_1a=m_2a\)
\((m_1+m_2)a=(m_2-m_1)g\)
\(a=\dfrac{(m_2-m_1)g}{m_1+m_2}\)
A \(6\,m/s^2\)
B \(4\,m/s^2\)
C \(8\,m/s^2\)
D \(3.33\,m/s^2\)
Free body diagrams
\(T-m_1g=m_1a\) ...(i)
\(m_2g-T=m_2a\) ...(ii)
\(\implies T =m_1g+m_1a\)
Putting value of T in equation (ii)
\(m_2g-m_1g-m_1a=m_2a\)
\((m_2-m_1)g=(m_1+m_2)a\)
\(a=\dfrac{(m_2-m_1)g}{m_1+m_2}\)
A \(\dfrac{5}{3}\,m/s^2,\dfrac{10}{3}\,m/s^2\)
B \(\dfrac{10}{3}m/s^2\;each\)
C \(\dfrac{5}{3}\,m/s^2\;each\)
D \(\dfrac{10}{3}\,m/s^2,\dfrac{5}{3}\,m/s^2\)
A \(6m/s^2\;up,\,\,\, 6m/s^2\;up,\,\,\,2m/s^2\;down\)
B \(\dfrac{40}{7}m/s^2\;up,\,\,\, \dfrac{60}{7}m/s^2\;down,\,\,\,\dfrac{30}{7}m/s^2\;up\)
C \(5m/s^2\;up, \,\,\,3m/s^2\;up,\,\,\,2m/s^2\;down\)
D \(\dfrac{10}{7}m/s^2\;up, \,\,\,\dfrac{10}{7}m/s^2\;up,\,\,\,\dfrac{30}{7}m/s^2\,down \)
\(m_1g\;sin\theta\;-T=m_1a\) ...(i)
\(T-m_2g=m_2a\) ...(ii)
\(\Rightarrow T=m_2(g+a)\)
Putting value of T in equation (i)
\(m_1g\;sin\theta\;-m_2g-m_2a=m_1a\)
\((m_1\;sin\theta\;-m_2)g=(m_1+m_2)a\)
\(a=\left(\dfrac{m_1sin\theta-m_2}{m_1+m_2}\right)g\)
A \(3\,m/s^2\)
B \(\dfrac{1}{7}\,m/s^2\)
C \(3.33\,m/s^2\)
D \(\dfrac{5}{3}\,m/s^2\)
Free body diagrams
\(T-m_1g\;sin\theta=m_1a\) ...(i)
\(m_2g\;sin\theta-T=m_2a\)
\(\implies T=m_2g\;sin\theta-m_2a\)
Putting value of T in equation (i)
\(m_2g\;sin\theta\;-m_2a-m_1g\;sin\theta=m_1a\)
\(m_2g\;sin\theta\;-m_1g\;sin\theta=m_1a+m_2a\)
\(\dfrac{(m_2-m_1)g\;sin\theta}{m_1+m_2}=a\)
A \(2\,m/sec^2\)
B \(3\,m/sec^2\)
C \(4\,m/sec^2\)
D \(5\,m/sec^2\)