Question

Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 4.8 kg) portrayed in the figure below move on a frictionless surface and a force F = 45 N acts as shown on the 4.8-kg block.
(a) Determine the acceleration given this system.

(b) Determine the tension in the cord connecting the 4.8-kg block and the 1.0-kg blocks.

(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block.

Answer 1

Let

$a \ be\ common\ acceleration$

$T_1\ be \ tension\ in \ cord\ connecting\ m_3\ and\ m_1$

And $T_2\ be \ between\ m_1\ and\ m_2$

force balancing for $m_3:$

$45-T_1=m_3a=4.8a\ \ \ \ .........(1)$

Balancing force for $m_1:$

$T_1-T_2=m_1a=a\ \ \ \ ..........(2)$

Balancing force for $m_3:$

$T_2=m_2a=2a\ \ \ \ \ .........(3)$

Equating the value of $T_2$ in $(2)$ and after that adding $(1)$ and $(2)$

We get

$45=7.8a\\Or\\a=5.77 \ m\ s^{-2}$

Putting value of $a\ in\ (3)$

We get

$T_2=2\times 5.77=11.54\ N$

Putting the value of $T_2$ in $(2)$

We get

$T_1=17.31\ N$

(a) accelration = $5.77\ m\ s^{-2}$

(b) tension in the cord connecting the 4.8-kg block and the 1.0-kg blocks-

$T_1=17.31\ N$

(c) force exerted by the 1.0-kg block on the 2.0-kg block

$T_2=11.54\ N$