Question

21. Figure shows three blocks attached by cords that
loop over frictionless pulleys. Block B lies on a
frictionless table; the masses are ma = 6.00 kg.
mg = 8.00 kg, and mc = 10.0 kg. When the blocks are
released, what is the tension in the cord at the right?
Please answer quickly
​

Answer 1

Given :

▪ Mass of block A = 6kg

▪ Mass of block B = 8kg

▪ Mass of block C = 10kg

To Find :

▪ Tension in the rightward string.

Concept :

➡ This question is completely based on the concept of Newton's second law of motion.

➡ As per this law, Net force acting on the system is defined as the product of mass and acceleration.

➡ First we have to find out acceleration of the system after that we can calculate tension in the string by the FBD method.

Calculation :

✴ Acceleration of the system :

→ Fnet = M × a

→ (m_c - m_a)g = Ma

→ (10-6) × 10 = (6+8+10)a

→ 40 = 24a

→ a = 1.67m/s²

✴ Net force on block C :

→ (m_c)g - T_1 = (m_c)a

→ (10×10) - T_1 = (10×1.67)

→ 100 - T_1 = 16.7

→ T_1 = 100 - 16.7

→ T_1 = 83.3N ✔

Answer 2

Given ,

Mass of block A = 6 kg

Mass of block B = 8 kg

Mass of block C = 10 kg

We need to find the tension in the rightward string.

Answer will be easy when we draw Free Body Diagrams . i.e., All the forces acting on the system.

A/s to Newton's Second Law , " Force is defined as the product of mass and acceleration "

$\implies \bold{F_{net}=Ma}\\\\\implies \bold{(m_c-m_a)g=Ma}\\\\\implies \bold{(10-6)10=(6+8+10)a}\\\\\implies \bold{a=\frac{40}{24}}\\\\\implies \bold{a=1.67\ m/s^2}$

Now we need to find tension on block C.

So for that note down all forces acting on that block. i.e., Gravitational force (Downwards) , Tension ( Upwards ) and net force is downwards .

$\underbrace{\bold{Net\ Force\ Acting\ on\ block\ C \; :}}\\\\\implies \bold{F_{net}=F_g-T}\\\\\implies \bold{m_c.a=m_c.g-T}\\\\\implies \bold{T=m_c.g-m_c.a}\\\\\implies \bold{T=10*10-10*1.67}\\\\\implies \bold{T=100-16.7}\\\\\implies \bold{\bf{\blue{T=83.3\ Newton}}}$

So Tension acted to the card is 83.3 Newtons.