Chapter 5, Problem 5.92AP

In Figure P5.46, the pulleys and pulleys the cord are light, all surfaces are frictionless, and the cord does not stretch. (a) How does the acceleration of block 1 compare with the acceleration of block 2? Explain your reasoning. (b) The mass of block 2 is 1.30 kg. Find its acceleration as it depends on the mass m₁ of block 1. (c) What If? What does the result of part (b) predict if m₁ is very much less than 1.30 kg? (d) What docs the result of part (b) predict if m₂ approaches infinity? (e) In this last case, what is the tension in the cord? (f) Could you anticipate the answers to parts (c), (d), and (e) without first doing part (b)? Explain.

Figure P5.46

To determine

The way in which the acceleration of block $1$ can be compared with the acceleration of block $2$.

Explanation of Solution

The acceleration of any object is defined as the rate of change of the velocity on an object with respect to time.

$1\text{cm}$ downward motion of block $2$ requires block $1$ to move $2\text{cm}$ forward. It means block $1$ always has twice the speed of block $2$ thus the acceleration of block $1$ is twice the of block $2$.

Then, the acceleration of block $1$ is given as, $a_1=2a_2$

Here, $a_1$ is the acceleration of block $1$ and $a_2$ is the acceleration of block $2$.

Conclusion:

Therefore, the acceleration of block $1$ is twice the of block $2$.

To determine

The acceleration of block $2$.

Answer to Problem 5.92AP

The acceleration of block $2$ is $\frac{\left(12.7\text{N}\right)}{4m_1+\left(1.30\text{kg}\right)}$.

Explanation of Solution

The mass of block $2$ is $1.30\text{kg}$.

Write the expression for the net force on block $1$ in the x-direction

  $\begin{array}{c}{\displaystyle \sum F_{\text{x}}}=m_1{a}_1\\ T=m_1{a}_1\end{array}$

Here, ${\displaystyle \sum F_{\text{x}}}$ is the net force on block $1$ in the x-direction, $T$ is the tension in the rope and $m_1$ is the mass of block $1$.

Substitute $2a_2$ for $a_1$ in above expression.

  $T=m_1\left(2a_2\right)$                                                       (I)

Write the expression for the net force on block $2$ in the x-direction

  $\begin{array}{c}{\displaystyle \sum F_{\text{y}}}=m_2{a}_2\\ T+T-m_{2}g=m_2\left(-a_2\right)\end{array}$

Here, ${\displaystyle \sum F_{\text{y}}}$ is the net force in the y-direction, $g$ is the acceleration due to gravity and $m_2$ is the mass of block $2$.

Rearrange the above expression

  $2T-m_{2}g=-m_2{a}_2$                                            (II)

Solve the equation (I) and equation (II).

  $\begin{array}{c}2\left(m_1\left(2a_2\right)\right)-m_{2}g=-m_2{a}_2\\ 4m_1{a}_2+m_2{a}_2=m_{2}g\\ a_2=\frac{m_{2}g}{4m_1+m_2}\end{array}$                                                           (III)

Substitute $1.30\text{kg}$ for $m_2$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above expression.

  $\begin{array}{c}{a}_2=\frac{\left(1.30\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)}{4m_1+\left(1.30\text{kg}\right)}\\ =\frac{\left(12.7\text{N}\right)}{4m_1+\left(1.30\text{kg}\right)}\end{array}$                                                             (IV)

Conclusion:

Therefore, the acceleration of block $2$ is $\frac{\left(12.7\text{N}\right)}{4m_1+\left(1.30\text{kg}\right)}$.

To determine

The result of part (b) if $m_1$ is very much lesser than $m_2$.

Answer to Problem 5.92AP

The acceleration of block $2$ when $m_1$ is very much lesser than $m_2$ is $9.8\text{m}/{\text{s}}^2$ down.

Explanation of Solution

When mass $m_1$ is very much less than $m_2$ then it can be neglected.

Rewrite equation (III)

  $\begin{array}{c}{a}_2=\frac{m_{2}g}{4m_1+m_2}\\ =\frac{m_2}{m_2}\left(\frac{g}{4\frac{m_1}{m_2}+1}\right)\end{array}$

Neglect $\frac{m_1}{m_2}$ and rewrite the above equation

  $\begin{array}{c}{a}_2=g\\ =9.8\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Substitute $9.8\text{}{\text{ m/s}}^2$ for $g$ in the above equation

  $a_2=9.8\text{m}/{\text{s}}^2$

Therefore, the acceleration for $m_1$ is very much less than $m_2$ is $9.8\text{m}/{\text{s}}^2$ down.

To determine

The result of part (b) when $m_1$ approaches infinity.

Answer to Problem 5.92AP

The acceleration for $m_1$ approaching infinity is $0\text{m}/{\text{s}}^2$.

Explanation of Solution

When the $m_1$ approaches infinity, then the equation (IV) is given as,

  $\begin{array}{c}{a}_2=\frac{\left(12.7\text{N}\right)}{4\left(\infty \right)+\left(1.30\text{kg}\right)}\\ =0\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the acceleration for $m_1$ approaches infinity is $0\text{m}/{\text{s}}^2$.

To determine

The tension in the cord when $m_1$ approaches infinity.

Answer to Problem 5.92AP

The tension in the cord when $m_1$ approaches infinity is $6.37\text{N}$.

Explanation of Solution

Recall the equation (II).

    $2T-m_{2}g=-m_2{a}_2$

Rearrange for $T$

  $T=\frac{m_{2}g-m_2{a}_2}{2}$

Conclusion:

Substitute $0\text{m}/{\text{s}}^2$ for $a_2$ to, $1.30\text{kg}$ for $m_2$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation to find $T$.

  $\begin{array}{c}T=\frac{\left(1.30\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)-\left(1.30\text{kg}\right)\left(0\text{m}/{\text{s}}^2\right)}{2}\\ =\frac{\left(1.30\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)}{2}\\ =6.37\text{N}\end{array}$

Therefore, the tension in the cord when $m_1$ approaches infinity is $6.37\text{N}$.

To determine

The possibility of predicting the answer to (c), part (d) and part (e) without doing part (b).

Explanation of Solution

As the mass of the bock $1$ approaches zero, the block $2$ almost undergoes free fall and hence the acceleration of block $2$ is same as that of the gravitational acceleration.

In the case, when the mass of block $1$ becomes infinite, the system is almost in equilibrium and hence there is no acceleration. Since there is no acceleration, the weight of the block $1$ is balanced by the tensional force along the cord. From this, the tension can be found.

Conclusion:

Therefore, Yes the answers to the part (c), part (d) and part (e) can be predicted without first doing part (b).