Question

(ii) Two masses 26 kg and 24 kg are attached to the ends of a string which passes
over a frictionless pulley. 26 kg is lying over a smooth horizontal table. 24 kg
mass is moving vertically downward. Find the tension in the string and the
acceleration in the bodies.

Answer 1

TOPIC :- Law's of motion

$\maltese\:\underline{\textsf{\textbf{AnsWer :}}}\:\maltese$

⏭ First of all make an free body diagram "FBD" for the given question [ I've attached the figure refer it ].

✤ Points to keep in mind while drawing FBD's :

✏ Assume all the objects to be point particle.

✏ Weight 'mg' acting vertically downward in direction.

✏Keep all the forces tail to tail.

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✪ CALCULATION ✪

$\longrightarrow\:\:\sf Acceleration= \dfrac{F_{net}}{M_{total}}$

$\longrightarrow\:\:\sf Acceleration= \dfrac{mg}{M_{total}}$

$\longrightarrow\:\:\sf Acceleration= \dfrac{mg}{m_1 + m_2}$

$\longrightarrow\:\:\sf Acceleration= \dfrac{(24)g}{24 + 26}$

$\longrightarrow\:\:\sf Acceleration= \dfrac{24 \times 10}{24 + 26}$

$\longrightarrow\:\:\sf Acceleration= \dfrac{240}{50}$

$\longrightarrow\:\: \underline{ \boxed{ \sf Acceleration= 4.8 \: {ms}^{ - 2} }}$

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✒Now, let's calculate the tension of the string. Refer figure No.2.

✒Here Acceleration (a) is acting vertically downward. So, dominating force will be mg.

✒By using Newton's second law of motion we have :

$\longrightarrow\:\:\sf m_1g - T = m_1a$

$\longrightarrow\:\:\sf 24 \times 10 - T = 24 \times 4.8$

$\longrightarrow\:\:\sf 240 - T = 115.2$

$\longrightarrow\:\:\sf T = 240 - 115.2$

$\longrightarrow\:\: \underline{ \boxed{\sf T = 124.8 \: N}}$

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Here, I've given step by step explaination. You can calculate the tension in the string and Acceleration by given below formula directly :

☢ Tension in the string :-

$\dag\:\bf T = \dfrac{m_1m_2}{m_1 + m_2}g$

☢ Acceleration :-

$\dag\:\bf a = \dfrac{m_1}{m_1 + m_2}g$

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Answer 2

Question :

(ii) Two masses 26 kg and 24 kg are attached to the ends of a string which passes

over a frictionless pulley. 26 kg is lying over a smooth horizontal table. 24 kg

mass is moving vertically downward. Find the tension in the string and the

acceleration in the bodies.

Let's Start :

• At first we start doing the calculation of acceleration

$\: \: \: \rightarrow\sf \: Acceleration = \large \frac{Fɴᴇᴛ}{Mᴛᴏᴛᴀʟ}$

$\: \: \: \rightarrow \sf\: Acceleration = \large\frac{mg}{Mᴛᴏᴛᴀʟ}$

$\: \: \: \rightarrow \sf\: Acceleration = \large \frac{mg}{m 1m2}$

$\: \: \: \rightarrow \sf \: Acceleration = \large\frac{(24)g}{24 + 26}$

$\: \: \: \rightarrow \sf \: Acceleration = \large \frac{24 \times 10}{24 + 26}$

$\: \: \: \tt \rightarrow \: Acceleration = \large\frac{240}{50}$

$\: \: \: \: \: \: { \underline{ \underline {\sf \: Acceleration =4.8m {s}^{ - 2} }}}$

Now Let's do the tension of the String :

$\: \: \: \: \implies \sf \: m1g - T = m1a$

$\: \: \: \: \: \:\rightarrow \sf24×10-T=24×4.8$

$\: \: \: \: \: \: \implies\sf240-T=115.2$

$\: \: \: \: \: \: \implies\sf T=240-115.2$

$\: \: \: \: \: \: \: \implies\sf T=124.8N$

Other Information :

Some Formula's :

$\: \: \: \: \: \implies \pink{\sf \: T = \large\frac{m1m2}{m1 + m2} }$

$\: \: \: \: \: \: \implies \sf \red{a = \large \frac{m1}{m1 + m2} }$