Answer:
The
free body diagrams of three blocks of masses M, \[{{m}_{1}}\] and \[{{m}_{2}}\]
are shown in Fig. 3(HT).4.
Equations of motion of three blocks are
Ma = F - T ???..
(i)
\[{{m}_{1}}{{a}_{1}}=2T-{{m}_{1}}g\] ???..
(ii)
\[{{m}_{2}}{{a}_{2}}={{m}_{2}}g-T\] ???..
(iii)
Now, if mass M moves to the left
through a distance x, and mass \[{{m}_{2}}\] moves downwards through the same distance
x, then the distance travelled by mass \[{{m}_{1}}\]is 2 x, upwards. Therefore,
sum of the accelerations of M
and \[{{m}_{2}}\] is double the
acceleration of \[{{m}_{1}}\] i.e.,
\[a+{{a}_{2}}=2{{a}_{1}}\]
...(iv)
From (i), \[T=F-Ma=20-2a\]
...(v)
From (ii), \[T={{m}_{2}}g-{{m}_{2}}{{a}_{2}}=2\times
9.8-2{{a}_{2}}\] ...(vi)
From (iii), \[2T=\,{{m}_{1}}{{a}_{1}}+{{m}_{1}}g=1{{a}_{1}}+9.8\]
...(vii)
Add (v) and (vi), \[2T=\left(
20-2a \right)+\left( 19.6-2{{a}_{2}} \right)=39.6-2\left( a+{{a}_{2}}
\right)=39.6-4{{a}_{1}}\]
Using (vii), \[{{a}_{1}}+9.8=39.6-4{{a}_{1}}\]
\[5{{a}_{1}}=39.6-9.8=29.8\]
\[{{a}_{1}}=\frac{29.8}{5}=5.96\,m/{{s}^{2}}\]
From (vi) 2 T = 5.96 + 9.8 = 15.76
From (vii), 2T = 19.6 - 7= 19.6 - 7.88 =
11.78
\[T=\frac{15.76}{2}\]
From(vi) \[2{{a}_{2}}=19.6-T=19.6-7.88=11.78\]
\[{{a}_{2}}=\frac{11.78}{2}=5.89\,m/{{s}^{2}}\]
From (iv) \[a=2{{a}_{1}}-{{a}_{2}}=2\times
5.96-5.89=11.92-5.89=6.03\,m/{{s}^{2}}\]
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