91Ó°ÊÓ

The pull for the system of sleds is $P=190\mathrm{N}$

The mass of sled A is $m_a=10\mathrm{kg}$

The mass of sled B is $m_b=20\mathrm{kg}$

The mass of sled C is $m_c=30\mathrm{kg}$

The acceleration of the systems is calculated by considering all the sleds as a single sled of total mass of sleds.

The expression for the force is given by,

$F=ma$

Here$m$is the mass, $F$ is the force and $a$ is the acceleration.

(a)

The acceleration of the system is calculated as:

$P=\left(m_a+m_b+m_c\right)a$

Here, $P$is the pull for sleds, $a$is the acceleration of the system of sleds.

Substitute 190N for P , 10kg for $m_a$, 20kg for $m_b$ and 30kg for $m_c$.

$$\begin{gathered} 190\mathrm{N}=\left(10\mathrm{kg}+20\mathrm{kg}+30\mathrm{kg}\right)\mathrm{a} \\ \mathrm{a}=3.2\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the system is $3.2\mathrm{m}/{\mathrm{s}}^2$.

(b)

The tension in the rope A is calculated as:

$P-T_a=m_{a}a$

Here, $T_a$ is the tension in rope A.

Substitute 190N for P, 10kg for $m_a$ and $3.2\mathrm{m}/{\mathrm{s}}^2$ for a.

$$\begin{gathered} 190\mathrm{N}-{\mathrm{T}}_{\mathrm{a}}=\left(10\mathrm{kg}\right)\left(3.2\mathrm{m}/{\mathrm{s}}^2\right) \\ {\mathrm{T}}_{\mathrm{a}}=158\mathrm{N} \end{gathered}$$

The tension in the rope B is calculated as:

$T_b=m_{c}a$

Here, $T_b$ is the tension in rope B.

Substitute 30kg for $m_c$ and $3.2\mathrm{m}/{\mathrm{s}}^2$ for a .

localid="1663742023525" $$\begin{gathered} T_b=\left(30\mathrm{kg}\right)\left(3.2\mathrm{m}/{\mathrm{s}}^2\right) \\ {T}_b=96\mathrm{N} \end{gathered}$$

Therefore, the tension in the ropes A and B are 158 N and 96 N .