Answer:
Let a be the common acceleration of
the whole system.
\[\therefore \] F = (m
+ m + m + m) a = 4 ma ; a = F/4 m
3. (a) 3.
(b)
Applying Newton's 2nd
law separately for each block, Fig. 3(a).24.
\[F-{{T}_{1}}=ma,\,{{T}_{1}}-{{T}_{2}}=ma;\,\,\,{{T}_{2}}-{{T}_{3}}=ma,\,{{T}_{3}}=ma\]
On solving these
equations, we get\[{{T}_{1}}=\frac{3}{4}F,\,\,\,{{T}_{2}}=\frac{1}{2}F,\,\,\,{{T}_{3}}=\frac{1}{4}F\]
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