Question
Find the acceleration of the blocks A and B in the three situations shown in figure.
✓
Answer
- $5\text{a + T}-5\text{g}=0\Rightarrow\text{T}=5\text{g}-5\text{a} \ ...(\text{i})$ (From FBD - 1)
From equn (i) and (ii), we get
$\text{5g}-5\text{a}=8\text{g}+16\text{a}\Rightarrow21\text{a}=-3\text{g}\Rightarrow\text{a}=-\frac{1}{7}\text{g}$
So, acceleration of 5kg mass is $\frac{\text{g}}{7}$ upward and that of 4kg mass is $2\text{a}=\frac{2\text{g}}{7}$ (downward).
$4\text{a}-\frac{\text{t}}{2}=0\Rightarrow8\text{a}-\text{T}=0\Rightarrow\text{T = 8a} \ ...(\text{ii})$ [from FBD - 4]
Again, $\text{T + 5a}-5\text{g}=0\Rightarrow8\text{a + 5a}-5\text{g}=0$
$\Rightarrow13\text{a}-5\text{g}=0\Rightarrow\text{a =}\frac{5\text{g}}{13}$ downward. (from FBD - 3)
Acceleration of mass
- $\text{kg is 2a}=\frac{10}{13}(\text{g}) \ \& \ 5\text{kg}$
- is $\frac{5\text{g}}{13}.$
$\text{T + 1a}-1\text{g}=0\Rightarrow\text{T = 1g}-1\text{a} \ ...(\text{i})$ [From FBD - 5]
Again, $\frac{\text{T}}{2}-2\text{g}-4\text{a}=0\Rightarrow\text{T}-4\text{g}-8\text{a}=0 \ ...(\text{ii})$ [From FBD - 6]
$\Rightarrow1\text{g}-1\text{a}-4\text{g}-8\text{a}=0$ [From (i)]
$\Rightarrow\text{a}=-\Big(\frac{\text{g}}{3}\Big)$ downward.
Acceleration of mass 1kg(b) is $\frac{\text{g}}{3}\text{(up)}$
Acceleration of mass 2kg(A) is $\frac{2\text{g}}{3}$ (downward).
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