Question

A man standing in a lift carrying a bag of 5 kg
If lift moves vertically downwards with acceleration
g/2. Find tension force in handle of bag.
​

Answer 1

Answer:

• The tension (T) in the bag is 25 N

Given:

1. Mass of bag (M) = 5 Kg.
2. acceleration (a) = g/2 m/s².

Explanation:

$\rule{300}{1.5}$

F.B.D of Bag is :-

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(10,6)\put(1,1){\line(1,0){2}}\put(1,1){\line(0,1){2}}\put(1,3){\line(1,0){2}}\put(3,1){\line(0,1){2}}\put(2,1){\vector(0,-1){2}}\put(2,3){\vector(0,1){2}}\put(5,4){\vector(0,-1){4}}\put(1.3,-1.5){ \sf Weight\;(Mg) }\put(1.4,5.2){ \sf Tension\;(T) }\put(5.2,2){ \sf acceleration\;(a) }\put(1.3,2){ \sf Mass\;(M) }\end{picture}$

Now, from F.B.D we can make out that,

$\large\bigstar\;\underline{\boxed{\sf F=Mg-T}}$

Here,

• F Denotes external force applied.
• Mg Denotes weight.
• T Denotes Tension.

Solving,

$\longrightarrow\sf F=Mg-T\\\\\\\longrightarrow\sf Ma=Mg-T\ \ \because\bigg(F=Ma\bigg)\\\\\\\longrightarrow\sf T=Mg-Ma\\\\\\\longrightarrow\sf T=M\;(g-a)$

Substituting the values,

$\longrightarrow\sf T=5\;\Bigg(g-\dfrac{g}{2}\Bigg)\\\\\\\longrightarrow\sf T=5\;\Bigg(\dfrac{2g-g}{2}\Bigg)\\\\\\\longrightarrow\sf T=5\times \dfrac{g}{2}\\\\\\\longrightarrow\sf T=5\times \dfrac{10}{2}\ \ \ \because\bigg(g=10\;m/s^{2}\bigg)\\\\\\\longrightarrow\sf T=5\times 5$

$\longrightarrow\sf T=25\\\\\\\longrightarrow \large{\underline{\boxed{\textsf{\textbf{T=25\;N}}}}}$

∴ The tension (T) in the bag is 25 N.

$\rule{300}{1.5}$