Chapter 5, Problem 80QAP

To determine

The magnitude of the acceleration of the block resting on an inclined plane.

Answer to Problem 80QAP

The magnitude of the acceleration of the block resting on an inclined plane is 3.77 m/s².

Explanation of Solution

Given info:

Mass of the block placed on the incline

  $m_1=1.00\text{ kg}$

Mass of the hanging block

  $m_2=2.00\text{ kg}$

Angle made by the plane with horizontal

  $\theta =30.0°$

Coefficient of static friction

  ${\mu }_s=0.500$

Coefficient of kinetic friction

  ${\mu }_k=0.400$

Formula used:

Free body diagrams are drawn for the two blocks and the acceleration of the block is determined using the force equations for both the blocks.

The free body diagram for the block of mass $m_2$, assuming the positive x axis pointing downwards and the y axis perpendicular to it is shown below.

Since m₂ is greater than m₁, the hanging block would tend to move down and the block on the incline would slide upwards along the incline.

The weight of the block is ${\overrightarrow{w}}_2$, which acts along the +x direction and the tension $\overrightarrow{T}$ acts along the −x direction.

  

The total force acting along the +x direction is given by,

  ${\displaystyle \sum F_{2x}}=w_2-T=m_2{a}_x$(1)

Here, a_x is the block s acceleration along the downward direction (+x ).

Draw the free body diagram for the block of mass $m_1$. Since the block tends to slide up the plane, the force of friction $\overrightarrow{f}$ acts downwards. The coordinate system under consideration has its +x direction up the incline and the +y direction perpendicular to the incline.

  

The weight ${\overrightarrow{w}}_1$ acts downwards and the normal force $\overrightarrow{n}$ acts along the +y direction. The force of friction $\overrightarrow{f}$ acts downwards along the −x direction and the tension $\overrightarrow{T}$ acts parallel to the plane along the +x direction.

Resolve the weight ${\overrightarrow{w}}_1$ into two components ${\overrightarrow{w}}_{1x}$ and ${\overrightarrow{w}}_{1y}$ along the −x and the +y directions.

Both the blocks have the same magnitude of acceleration.

The force equation along the +x direction is given by,

  ${\displaystyle \sum F_{1x}}=T-w_{1x}-f=m_1{a}_x$(2)

The force equation along the +y direction is given by,

  ${\displaystyle \sum F_{1y}}=n-w_{1y}$(3)

Since the block is in equilibrium in the y direction,

  ${\displaystyle \sum F_{1y}}=n-w_{1y}=0$

Hence,

  $n=w_{1y}$.....(4)

The force of friction and the normal force are related as follows:

  $f=\mu n$... (5)

The value of the coefficient of friction $\mu$ takes the value of ${\mu }_s$ if the block is at rest and ${\mu }_k$ if the block is in motion.

Calculation:

First determine, if the system is at rest or in motion.

The system will be at rest if

  $T<w_{1x}+f_s$

If the system is at rest, equation (1) can be written as,

  $\begin{array}{c}{\displaystyle \sum F_{2x}}=w_2-T=m_2{a}_x=0\\ T=w_2\end{array}$

Since $w_2=m_{2}g$, where g is the acceleration of free fall

  $T=m_{2}g$

Substitute the values of the variables in the above equation,

  $\begin{array}{c}T=m_{2}g\\ =\left(2.00\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\\ =19.6\text{ N}\end{array}$

Calculate the value of $w_{1x}+f_s$.

The component $w_{1x}=w_1\sin \theta$, where, $w_1=m_{1}g$.

  $\begin{array}{c}{w}_{1x}=w_1\sin \theta \\ =m_{1}g\sin \theta \\ =\left(1.00\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(\sin 30.0°\right)\\ =4.90\text{ N}\end{array}$

If the system is at rest, assume the maximum force of static friction to act on the block.

Then,

  $f_s={\mu }_{s}n$

From equation (4),

  $\begin{array}{c}{f}_s={\mu }_{s}n\\ ={\mu }_s{w}_{1y}\end{array}$

The component $w_{1y}$ is given by,

  $w_{1y}=w_1\cos \theta$, where $w_1=m_{1}g$

Calculate the maximum force of static friction acting on the block.

  $\begin{array}{c}{f}_s={\mu }_s{w}_{1y}\\ ={\mu }_s{m}_{1}g\cos \theta \\ =\left(0.500\right)\left(1.00\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(\cos 30.0°\right)\\ =4.24\text{ N}\end{array}$

Therefore,

  $\begin{array}{c}{w}_{1x}+f_s=\left(4.90\text{ N}\right)+\left(4.24\text{ N}\right)\\ =9.14\text{ N}\end{array}$

Since it is seen that $T>w_{1x}+f_s$, the system accelerates. The force of friction is that of kinetic friction $f_k$.

Add equations (1) $w_2-T=m_2{a}_x$ and (2) $T-w_{1x}-f_k=m_1{a}_x$

  $\begin{array}{c}T-w_{1x}-f_k=m_1{a}_x\\ w_2-w_{1x}-f_k=\left(m_1+m_2\right)a_x\end{array}$

Substitute the expressions for w₂,w_1x and f_k in the expression and write an expression for a_x.

  $m_{2}g-m_{1}g\sin \theta -{\mu }_k{m}_{1}g\cos \theta =\left(m_1+m_2\right)a_x$

  $a_x=\frac{m_{2}g-m_{1}g\sin \theta -{\mu }_k{m}_{1}g\cos \theta }{\left(m_1+m_2\right)}$

Substitute the values of the variables in the expression and determine the magnitude of the acceleration of the block on the incline.

  $\begin{array}{c}{a}_x=\frac{m_{2}g-m_{1}g\sin \theta -{\mu }_k{m}_{1}g\cos \theta }{\left(m_1+m_2\right)}\\ =\frac{\left(2.00\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)-\left(1.00\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(\sin 30.0°\right)-\left(0.400\right)\left(1.00\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(\cos 30.0°\right)}{\left(1.00\text{ kg}\right)+\left(2.00\text{ kg}\right)}\\ =3.77{\text{ m/s}}^2\end{array}$

Conclusion:

The magnitude of the acceleration of the block resting on an inclined plane is 3.77 m/s².