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# Find acceleration in the given cases.

Last updated date: 20th Jun 2024
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Hint: First write the equation of motion of each block. Take into consideration the tension produced in the string, weight (mg) component along the inclined plane.
TENSION: whenever any string is tight then the end of the string pulls whatever is connected to them in inward direction. This inward pulling force of the string on the object is called a tension force. It depends on mass, as mass increases tension also increases.
Remember that the heavy block will move down and the light block will slide up. According to this, it gives direction to the acceleration produced.

Complete step by step answer

Let T, be the tension produced in the string and ‘a’ be the acceleration produced in each block.
Therefore, equation of motion of block A can be written as
\begin{align} & {{m}_{A}}g\sin \theta -T={{m}_{A}}a \\ & {{m}_{A}}g\sin 37{}^\circ -T={{m}_{A}}a\text{ }............\text{(1)} \\ & \text{Equation of motion for block B is:-} \\ & T-{{m}_{B}}g\sin {{\theta }_{2}}={{m}_{B}}a\text{ } \\ & T-{{m}_{B}}g\sin 53{}^\circ ={{m}_{B}}a\text{ }.............\text{(2)} \\ \end{align}
Adding equations 1. and 2. Gives,
\begin{align} & g({{m}_{A}}\sin 37{}^\circ -{{m}_{b}}\sin 53{}^\circ )=({{m}_{A}}+{{m}_{B}})a \\ & g(100\times \dfrac{3}{5}-50\times \dfrac{4}{5})=(100+50)a \\ & 10(60-40)=150a \\ & a=\dfrac{200}{15}=\dfrac{4}{3}m/{{s}^{2}} \\ & \text{Thus acceleration of block= 4/3 ms}_{^{{}}}^{-2} \\ \end{align} .

\begin{align} & Here,\angle LMN=\angle BAC=\theta \\ & \\ \end{align}
$\text{As here, MN}\bot \text{AC and BA}\bot \text{AD}$
Thus component along AN = mg $\text{sin }\!\!\theta\!\!\text{ }$
Component along AB = mg $\text{cos }\!\!\theta\!\!\text{ }$ .