Q14E - University Physics with Modern Physics

Question

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.

           

Step-by-Step Solution

Answer

(a) The acceleration of the system is  $3.2 \mathrm{m}/{\mathrm{s}}^2$.

(b) The tensions in the ropes A and B are 158 N  and 96 N.

Step 1: Given Data:

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 $m_c=30 \mathrm{kg}$ 

Step 2: Acceleration of system

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.

Step 3: Determine the acceleration of the system.

(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 $3.2 \mathrm{m}/{\mathrm{s}}^2$ .

Step 4: Determine the tensions in rope A and B.

(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 .

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

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