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Question: Three blocks of masses \( {{\text{m}}_{1}},{{\text{m}}_{2}} \) and \( {{\text{m}}_{3}} \) are placed...

Three blocks of masses m1,m2{{\text{m}}_{1}},{{\text{m}}_{2}} and m3{{\text{m}}_{3}} are placed on a horizontal frictionless surface. A force of 40 N pulls the system. Then calculate the value of T if
m1=10 kg, m2=6 kg, m3=4 kg{{\text{m}}_{1}}=10\text{ kg, }{{\text{m}}_{2}}=6\text{ kg, }{{\text{m}}_{3}}=4\text{ kg}

(A) 40 N
(B) 20 N
(C) 10 N
(D) 5 N

Explanation

Solution

We know that tension T can be formed by using the formula
T=ma
Now, at the point A tension T is due to sum of masses m2{{\text{m}}_{2}} and m3{{\text{m}}_{3}}
So,
T=(m2+m3)a\text{T=}\left( {{\text{m}}_{2}}+{{\text{m}}_{3}} \right)\text{a}
Using this formula, tension T can be found.

Complete step by step solution
We know that F=ma
Here we have 3 masses, m1=10 kg, m2=6 kg, m3=4 kg{{\text{m}}_{1}}=10\text{ kg, }{{\text{m}}_{2}}=6\text{ kg, }{{\text{m}}_{3}}=4\text{ kg}
So F= (m1+m2+m3)a\left( {{\text{m}}_{1}}+{{\text{m}}_{2}}+{{\text{m}}_{3}} \right)\text{a}
=(10+6+4)a=\left( 10+6+4 \right)\text{a}
F=20 a
F=40 N…… (Given)
So, 40=20a
a=2
Now tension T is given as:
T= (m2+m3)a\left( {{\text{m}}_{2}}+{{\text{m}}_{3}} \right)\text{a}
=(6+4)×2 =10×2 \begin{aligned} & =\left( 6+4 \right)\times 2 \\\ & =10\times 2 \\\ \end{aligned}
T=20 N
So, the correct option is (B).

Note
Tension is described as the pulling force transmitted axially by means of a string, cable, chain or 1-D objects. Tension never applies on its own. It has to be put on a system and it is always a pulling force.