Question

Three cubical shaper bodies having masses m=20 kg, m2 40
kg and m=60 kg are lying on frictionless plane as shown in
the following figure. All the three bodies are connected by
inextensible strings, tensions Ti and T2 will be:
F-120 N​

Answer 1

Solution :

please refer the attachment first

Step I : Draw all the forces acting on the blocks

Step II : Find acceleration of the system .

• Force applied = 120 N
• Total mass = 20 + 40 + 60 = 120 kg

By applying Newton's second law of motion ,

⇒ F = ma

⇒ a = F / m

⇒ a = 120 / 120

⇒ a = 1 m/s²

Net force acting on 20 kg block

$\leadsto F - T _1= m_1a$

$\leadsto 120 - T _1= 20 \times 1$

$\leadsto T _1=100 N$

Net force acting on the 60 kg block

$\leadsto T _2 = m_3a$

$\leadsto T _2 = 60 \times 1$

$\leadsto T _2 = 60 N$

OR

Net force acting on the 40 kg block

$\leadsto T_1 - T _2 = m_2 a$

$\leadsto 100 - T _2 = 40 \times 1$

$\leadsto T _2 = 60 N$