Chapter 5, Problem 47PQ

A block with mass m₁ hangs from a rope that is extended over an ideal pulley and attached to a second block with mass m₂ that sits on a ledge. The second block is also connected to a third block with mass m₃ by a second rope that hangs over a second ideal pulley as shown in Figure P5.47. If the friction between the ledge and the second block is negligible, m₁ = 3.00 kg, m₂ = 5.00 kg, and m₃ = 8.00 kg, find the magnitude of the tension in each rope and the acceleration of each block.

FIGURE P5.47

To determine

What is the magnitude of tension on each rope and the acceleration on each block?

Answer to Problem 47PQ

The magnitude of tension on each rope is $38.6\text{ N}$ & $53.9\text{ N}$ and the acceleration on each block is $\text{3}{\text{.07 m/s}}^2$.

Explanation of Solution

From the given condition the $8.00\text{ kg}$ object will accelerate downwards, $3.00\text{ kg}$ object will accelerate upwards and $5.00\text{ kg}$ object will move to the right. As the three object connected together they all have same acceleration but the tension in each rope will vary. The free body diagram is given below.

Applying Newton’s laws.

For $m_1$:

    $\Sigma F_y=F_{T_1}-F_{g1}=m_{1}a$                                                                                (I)

Here, $F_y$ is the force, $F_{T_1}$ is the tension force due to $m_1$, $m_1$ is the mass, $F_{g1}$ is the gravitational force on $m_1$ and $a$ is the acceleration.

For $m_2$:

    $\Sigma F_x=F_{T_2}-F_{T_1}=m_{2}a$                                                                       (II)

  $\Sigma F_y=F_N-F_{g2}=0$                                                                                (III)

Here, $F_{g2}$ is the gravitational force on $m_2$ and $F_N$ is the normal force.

For $m_3$:

      $\Sigma F_y=F_{g3}-F_{T_2}=m_{3}a$                                                                             (IV)

Here, $F_{g3}$ is the gravitational force on $m_3$ and $F_{T_2}$ is the tension force due to $m_2$.

Write the equation for gravitational force.

    $F_g=mg$                                                                                              (V)

Here, $g$ is the acceleration due to gravity.

Conclusion:

Using equation V find the gravitational force on each object.

    $\begin{array}{c}{F}_{g1}=m_{1}g\\ =\left(3.00\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\\ =29.4\text{ N}\end{array}$

    $\begin{array}{c}{F}_{g2}=m_{1}g\\ =\left(5.00\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\\ =49.1\text{ N}\end{array}$

    $\begin{array}{c}{F}_{g3}=m_{3}g\\ =\left(8.00\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\\ =78.5\text{ N}\end{array}$

Substitute the above values in equation I, II and IV to generate another three equation with unknown factors.

    $F_{T_1}-29.4\text{ N}=\left(3.00\text{ kg}\right)a$                                                                                (VI)

    $F_{T_2}-F_{T_1}=\left(5.00\text{ kg}\right)a$                                                                                (VII)

  $78.5\text{ N}-F_{T_2}=\left(8.00\text{ kg}\right)a$                                                                             (VIII)

Solve the equation VI and VIII to get tension and substitute in equation VII.

    $F_{T_1}=\left(3.00\text{ kg}\right)a+29.4\text{ N}$                                                                            (IX)

    $F_{T_2}=78.5\text{ N}-\left(8.00\text{ kg}\right)a$                                                                            (X)

Then equation VII becomes,

    $\begin{array}{c}{F}_{T_2}-F_{T_1}=\left(5.00\text{ kg}\right)a\\ \left[78.5\text{ N}-\left(8.00\text{ kg}\right)a\right]-\left[\left(3.00\text{ kg}\right)a+29.4\text{ N}\right]=\left(5.00\text{ kg}\right)a\\ 49.1\text{ N}-\left(11.00\text{ kg}\right)a=\left(5.00\text{ kg}\right)a\\ a=\frac{49.1\text{ N}}{16.00\text{ kg}}\\ =\text{3}{\text{.07 m/s}}^2\end{array}$

Substitute $\text{3}{\text{.07 m/s}}^2$ for $a$ in equation IX and X to get the tension force.

    $\begin{array}{c}{F}_{T_1}=\left(3.00\text{ kg}\right)\left(\text{3}{\text{.07 m/s}}^2\right)+29.4\text{ N}\\ \text{=38}.6\text{ N}\end{array}$

    $\begin{array}{c}{F}_{T_2}=78.5\text{ N}-\left(8.00\text{ kg}\right)\left(\text{3}{\text{.07 m/s}}^2\right)\\ =53.9\text{ N}\end{array}$

Therefore, the magnitude of tension on each rope is $38.6\text{ N}$ & $53.9\text{ N}$ and the acceleration on each block is $\text{3}{\text{.07 m/s}}^2$.