Q48P - University Physics with Modern Physics

Question

The two blocks in Fig. P4.48 are connected by a heavy uniform rope with a mass of 4.00 kg . An upward force of 200 N is applied as shown. (a) Draw three free-body diagrams: one for the 6.00 kg block, one for the 4.00 kg rope, and another one for the 5.00 kg block. For each force, indicate what body exerts that force. (b) What is the acceleration of the system? (c) What is the tension at the top of the heavy rope? (d) What is the tension at the midpoint of the rope?

Step-by-Step Solution

Answer

     

(a) , .

For the first block, the upward force mainly exerts the force.

.

For the rope, the upward force of the block mainly exerts the force as that is higher than the second block.

.

For the second block, the upward force of the rope mainly exerts the force.

(b) The acceleration of the system is $3.53 \mathrm{m}/{\mathrm{s}}^2$.

(c) The tension at the top of the heavy rope is $120.02 \mathrm{N}$.

(d) The tension at the midpoint of the rope is $93.31 \mathrm{N}$.

Step 1: Identification of the given data

The given data can be listed below as:

• The mass of the first block, $M_1=6.00 \mathrm{kg}$
• The mass of the first block, $M_2=5.00 \mathrm{kg}$
• The mass of the rope is, $m=4.00 \mathrm{kg}$
• The force exerted in the upward direction is,  $F=200 \mathrm{N}$

Step 2: Significance of the acceleration of a system

The acceleration is directly proportional to the force and inversely proportional to the mass of that object. The acceleration of an object mainly helps to give the force of that object.

Step 3: (a) Determination of the free body diagrams and the body

The free body diagram of the block has been drawn below:

 

                                     

 

In the above diagram, a constant force $F$is exerted on the top of the block and the weight of the block is acting downwards. For the first block, the upward force mainly exerts the force.

 

The free body diagram of the $4.00 \mathrm{kg}$rope has been drawn below:

 

 

In the above diagram, a constant tension $T$ is exerted on the top of the rope and the weight of the rope is acting downwards. For the rope, the upward force of the block mainly exerts the force as that is higher than the second block.

 

The free body diagram of the $5.00 \mathrm{kg}$ block has been drawn below:

                                 

In the above diagram, a constant tension $T$ is exerted on the top of the block and the weight of the block $w$ is acting downwards. For the second block, the upward force of the rope mainly exerts the force.

Step 4: (b) Determination of the acceleration of the system

The equation of the acceleration of the system can be expressed as:

 

 $$\begin{gathered} F-\left(M_1+M_2+m\right)g=\left(M_1+M_2+m\right)a \\ a=\frac{F-\left(M_1+M_2+m\right)g}{\left(M_1+M_2+m\right)} \end{gathered}$$

 

Here, $a$ is the acceleration of the system, $F$ is the force exerted in the upward direction, $g$ is the acceleration due to gravity, $M_1$ is the mass of the first block,  $M_2$ is the mass of the second block and $m$ is the mass of the rope.

 

Substitute the values in the above equation. 

 $$\begin{gathered} \mathrm{a}=\frac{200 \mathrm{N}-\left(6.00 \mathrm{kg}+5.00 \mathrm{kg}+4.00 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)}{\left(6.00 \mathrm{kg}+5.00 \mathrm{kg}+4.00 \mathrm{kg}\right)} \\ =\frac{200 \mathrm{N}-\left(15.00 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)}{\left(15.00 \mathrm{kg}\right)} \\ =\frac{200 \mathrm{N}\times {\displaystyle \frac{1 \mathrm{kg}·{\mathrm{ms}}^2}{1\mathrm{N}}}\left(147 \mathrm{kg}· \mathrm{m}/{\mathrm{s}}^2\right)}{\left(15.00 \mathrm{kg}\right)} \\ =\frac{200 \mathrm{kg}·{\mathrm{ms}}^2\left(147 \mathrm{kg}· \mathrm{m}/{\mathrm{s}}^2\right)}{\left(15.00 \mathrm{kg}\right)} \end{gathered}$$

Hence, further as:

 $$\begin{gathered} \mathrm{a}=\frac{200 \mathrm{kg}·{\mathrm{ms}}^2\left(147 \mathrm{kg}· \mathrm{m}/{\mathrm{s}}^2\right)}{\left(15.00 \mathrm{kg}\right)} \\ =\frac{53 \mathrm{kg}·{\mathrm{ms}}^2}{\left(15.00 \mathrm{kg}\right)} \\ =3.53 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration of the system is $3.53 \mathrm{m}/{\mathrm{s}}^2$.

Step 5: (c) Determination of the tension at the top of the rope

The equation of the tension at the top of the rope is expressed as:

 $$\begin{gathered} F-T=M_{1}g+M_{1}a \\ T=F-M_1\left(g+a\right) \end{gathered}$$

Here, $T$ is the tension at the top of the rope.

 

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{T}=200 \mathrm{N}-6.00 \mathrm{kg}\left(9.8 \mathrm{m}/{\mathrm{s}}^2+3.53 \mathrm{m}/{\mathrm{s}}^2\right) \\ =200 \mathrm{N}-6.00 \mathrm{kg}\times 13.33\mathrm{m}/{\mathrm{s}}^2 \\ =200 \mathrm{N}-79.98 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =120.02 \mathrm{N} \end{gathered}$$ 

 

Thus, the tension at the top of the heavy rope is $120.02 \mathrm{N}$.

Step 6: (d) Determination of the tension at the midpoint

The equation of the tension at the midpoint is expressed as:

 

 $$\begin{gathered} T-\left(M_2+\frac{m}{2}\right)g=\left(M_2+\frac{m}{2}\right)a \\ T=\left(M_2+\frac{m}{2}\right)g+\left(M_2+\frac{m}{2}\right)a \end{gathered}$$

 

Here, $T$ is the tension at the midpoint of the rope.

 

Substitute the values in the above equation.

 

$$\begin{gathered} T=\left(5.00 \mathrm{kg}+\frac{4.00 \mathrm{kg}}{2}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)+\left(5.00 \mathrm{kg}+\frac{4.00 \mathrm{kg}}{2}\right)\left(3.53 \mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(5.00 \mathrm{kg}+2.00 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)+\left(5.00 \mathrm{kg}+2.00 \mathrm{kg}\right)\left(3.53 \mathrm{m}/{\mathrm{s}}^2\right) \\ =7.00 \mathrm{kg}\left(9.8 \mathrm{m}/{\mathrm{s}}^2+3.53 \mathrm{m}/{\mathrm{s}}^2\right) \\ =7.00 \mathrm{kg}\left(13.33\mathrm{m}/{\mathrm{s}}^2\right) \end{gathered}$$

 

Hence, further as:

 

$$\begin{gathered} \mathrm{T}=7.00 \mathrm{kg}\left(13.33\mathrm{m}/{\mathrm{s}}^2\right) \\ =93.31 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =93.31 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =93.31\mathrm{N} \end{gathered}$$

 Thus, the tension at the midpoint of the rope is $93.31\mathrm{N}$.