Chapter 4, Problem 84AP

In Figure P1.84, the pulleys and 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 m₂ = 1.30 kg. Derive an expression for the acceleration of the block having mass m₂ as a function of the mass of block 1, m₁. (c) What does the result of part (b) predict if m₁ is very much less than 1.30 kg? (d) What does the result of part (b) predict if m₁ 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 P1.84

To determine

To compare the acceleration of the block 1 and block 2.

Answer to Problem 84AP

The forces of tension on the cord and the gravitational force on the block cause the acceleration of the two blocks.

Explanation of Solution

The two blocks accelerated towards the direction of its corresponding forces. The block 1 is accelerated in the direction of the tension in the cord, which is attached by the block1. Similarly, the block is accelerated downwards in the direction of gravitational force.

In this case if block 2 moves, say 1 unit downward then the cord attached in horizontal table would move 2 units because of the addition of tension on the cord above the block 2. Therefore the block 1 would have twice the velocity of that of block 1 and hence the acceleration too.

Conclusion:

Therefore, the acceleration of the block 1 is twice as that of block 2.

To determine

The magnitude of the acceleration of the mass $m_2$.

Answer to Problem 84AP

Solution:

The magnitude of the acceleration of the mass $m_2$ is $\frac{12.7\text{N}}{\left(4m_1+1.30\text{kg}\right)}$.

Explanation of Solution

Given Info:

Mass of block 2 is $1.30\text{kg}$ and acceleration due to gravity is $9.80\text{m}/{\text{s}}^{\text{2}}$.

Apply the Newton’s second law for the block 1.

$T=m_1{a}_1$ (I)

• T is the tension on the string
• $m_1$ is the mass of the block 1
• $a_1$ is the acceleration of the block 1

Since the acceleration of the block 1 is twice as that of block 2.

Write the formula to calculate $a_1$.

$a_1=2a_2$ (II)

• $a_2$ is the acceleration of the block 2

Rewrite the equation (I) using (II).

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

Apply Newton’s second law for the block 2.

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

• $m_2$ is the mass of the block 2
• g is the acceleration due to gravity

Substitute (III) in (IV) to calculate $a_2$.

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

Substitute $1.30\text{kg}$ for $m_2$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g in (V) to calculate $a_2$.

$\begin{array}{c}{a}_2=\frac{\left(1.30\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)}{\left(4m_1+1.30\text{kg}\right)}\\ =\frac{12.7\text{N}}{\left(4m_1+1.30\text{kg}\right)}\end{array}$

Conclusion:

Therefore, the magnitude of the acceleration of the mass $m_2$ is $\frac{12.7\text{N}}{\left(4m_1+1.30\text{kg}\right)}$.

To determine

To determine: The magnitude of the acceleration of the mass $m_2$ if mass of the block 1 is very small.

Answer to Problem 84AP

Solution:

The magnitude of the acceleration of the mass $m_2$ if mass $m_1$ too small is $9.77\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Explanation:

Given Info:

If the mass of the block 1 is very small compared to the mass of block 2 then mass of the block 1 is taken as zero.

Write the formula to calculate the acceleration of the block 2.

$a_2=\frac{12.7\text{N}}{\left(4m_1+1.30\text{kg}\right)}$

Substitute 0 for $m_1$ to calculate $a_2$.

$\begin{array}{c}{a}_2=\frac{12.7\text{N}}{\left(4\left(0\right)+1.30\text{kg}\right)}\\ =\frac{12.7\text{N}}{1.30\text{kg}}\\ =9.77\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the magnitude of the acceleration of the mass $m_2$ if mass $m_1$ too small is $9.77\text{m}/{\text{s}}^{\text{2}}$.

To determine

To determine: The magnitude of the acceleration of the mass $m_2$ if mass of the block 1 is infinity.

Answer to Problem 84AP

Solution:

The magnitude of the acceleration of the mass $m_2$ if mass $m_1$ is infinity is $0\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Explanation:

Given Info:

The mass of the block 1 $m_1$ is infinity.

Write the formula to calculate the acceleration of the block 2.

$a_2=\frac{12.7\text{N}}{\left(4m_1+1.30\text{kg}\right)}$

Substitute $\infty$ for $m_1$ to calculate $a_2$.

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

Conclusion:

Therefore, the magnitude of the acceleration of the mass $m_2$ if mass $m_1$ infinity is $0\text{m}/{\text{s}}^{\text{2}}$.

To determine

To determine: The magnitude of the tension in the cord if the mass of the block 1 is infinity.

Answer to Problem 84AP

Solution:

The magnitude of the tension in the cord if mass $m_1$ is infinity is $6.35\text{N}$.

Explanation of Solution

Given Info:

The mass of the block 1 $m_1$ is infinity, the mass of the block 2 is $1.30\text{kg}$  and the acceleration of block 2 is $0\text{m}/{\text{s}}^{\text{2}}$.

Write the Newton’s second law for the block 2 from (IV)

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

Substitute $1.30\text{kg}$ for $m_2$, $0\text{m}/{\text{s}}^{\text{2}}$ for $a_2$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate T.

$\begin{array}{c}\left(1.30\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)-2T=\left(1.30\text{kg}\right)\left(0\text{m}/{\text{s}}^{\text{2}}\right)\\ 12.7\text{N-2}T=0\\ 2T=12.7\text{N}\\ T=\frac{12.7\text{N}}{2}\\ =6.35\text{N}\end{array}$

Conclusion:

Therefore, the magnitude of the tension in the cord if mass $m_1$ is infinity is $6.35\text{N}$.

To determine

To determine: Is it possible to work part (c), (d) and (e) without calculating the expression in (b).

Answer to Problem 84AP

Answer:

It is possible to find the answers for whole parts.

Explanation of Solution

Explanation:

It is possible to calculate the whole parts without calculating the expression for the acceleration of the block 2. For the first case if the mass of the block 1 tends to zero, the block 2 would encounter a free fall. Therefore the acceleration of the block 2 is essentially zero.

In the second case as the mass of the block 1 is tends to infinity then for the block 2 it is impossible to move downward which would results $a_2\approx 0\text{m}/{\text{s}}^{\text{2}}$. The last one is the tension on the cord if mass of the block 1 tends to infinity. In that case the block 2 is supported by the tension on the two cords equally distributed. Therefore the tension essentially be the half of the weight of the block 2.

Conclusion:

Therefore, it is possible to find the answers for whole parts