Question
Two masses of $5kg$ and $3kg$ are suspended with help of massless inextensible strings as shown in Calculate $T_1$ and $T_2$ when whole system is going upwards with acceleration = $2m s^2$(use $g = 9.8m s^{–2}$).
✓
Answer
As the whole system is going up with acceleration = $a = 2ms-2\text{m}_1=5\text{kg}\ \text{m}_2=3\text{kg}\ \text{g}=9.8\text{m/ s}^2$
Tension in a string is equal and opposite in all parts of a string.
Forces on mass $m_1$
$\text{T}_1-\text{T}_2-\text{m}_1\text{g}=\text{m}_1\text{a}$
$\text{T}_1-\text{T}_2-5\text{g}+5\text{a}$
$\text{T}_1-\text{T}_2=5\text{g}+5\text{a}$
$\text{T}_1-\text{T}_2=5(9.8+2)$
$=5\times11.8$
$\text{T}_1-\text{T}_2=59.0\ \text{N}$
Forces on mass $m_2$
$\text{T}_2-\text{m}_2\text{g}=\text{m}_2\text{a}$
$\text{T}_2=\text{m}(\text{g}+\text{a})=3(9.8+2)=3\times11.8$
$\text{T}_2=35.4$
$\text{T}_1=\text{T}_2+59.0\Rightarrow\text{T}_1=53.4+59.0=94.4\ \text{N}$
Tension in a string is equal and opposite in all parts of a string.
Forces on mass $m_1$
$\text{T}_1-\text{T}_2-\text{m}_1\text{g}=\text{m}_1\text{a}$
$\text{T}_1-\text{T}_2-5\text{g}+5\text{a}$
$\text{T}_1-\text{T}_2=5\text{g}+5\text{a}$
$\text{T}_1-\text{T}_2=5(9.8+2)$
$=5\times11.8$
$\text{T}_1-\text{T}_2=59.0\ \text{N}$
Forces on mass $m_2$
$\text{T}_2-\text{m}_2\text{g}=\text{m}_2\text{a}$
$\text{T}_2=\text{m}(\text{g}+\text{a})=3(9.8+2)=3\times11.8$
$\text{T}_2=35.4$
$\text{T}_1=\text{T}_2+59.0\Rightarrow\text{T}_1=53.4+59.0=94.4\ \text{N}$
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