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

1.	(ii) Two masses 26 kg and 24 kg are attached to the ends of a string which passesover 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» 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. $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$ ✪ CALCULATION ✪ $\longrightarrow\:\:\sf Acceleration= \dfrac{F_{net}}{M_{total}} \\$ $\longrightarrow\:\:\sf Acceleration= \dfrac{mg}{M_{total}} \\$ $\longrightarrow\:\:\sf Acceleration= \dfrac{mg}{m_1 + m_2} \\$ $\longrightarrow\:\:\sf Acceleration= \dfrac{(24)g}{24 + 26} \\$ $\longrightarrow\:\:\sf Acceleration= \dfrac{24 \times 10}{24 + 26} \\$ $\longrightarrow\:\:\sf Acceleration= \dfrac{240}{50} \\$ $\longrightarrow\:\: \underline{ \boxed{ \sf Acceleration= 4.8 \: {ms}^{ - 2} }} \\$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$ ✒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 : $\longrightarrow\:\:\sf m_1g - T = m_1a \\$ $\longrightarrow\:\:\sf 24 \times 10 - T = 24 \times 4.8 \\$ $\longrightarrow\:\:\sf 240 - T = 115.2 \\$ $\longrightarrow\:\:\sf T = 240 - 115.2 \\$ $\longrightarrow\:\: \underline{ \boxed{\sf T = 124.8 \: N}}\\$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$ 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 :- $\dag\:\bf a = \dfrac{m_1}{m_1 + m_2}g$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

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

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.

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

✪ CALCULATION ✪

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

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

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

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

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

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

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

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

✒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 :

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

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

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

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

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

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

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 :-

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

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$