Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal ropes (Fig. E5.14). The pull is of magnitude 190N. Find (a) the acceleration of the system and (b) the tension in ropes A and B.

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 .