Chapter 5, Problem 42P

To determine

To evaluate: The acceleration of the two blocks sliding down is to be evaluated.

Answer to Problem 42P

The acceleration of each block is 0.758 m/s²

Explanation of Solution

Introduction:Newton's second law explains that the acceleration of an object is dependent upon two variables. Those are net force acting on the object and the object’s mass.

When the blocks are going down the plane. The frictional force should be upward the plane.

It is known that the component of the gravitational force down the plane is mg cosθ and the

component perpendicular to the plane is mg sin θ as shown in the figure below.

  

The frictional force acting on the first block is,

  $f1=\mu \text{1}FN1$

Here, $\mu \text{1}$ is coefficient of kinetic friction between the first block and the incline and $FN1$

normal force on the first block.

Applying Newton's second law to the first block, the net force along the horizontal direction is

  $\begin{array}{l}{\displaystyle \sum Fx}=m1gsin\theta +T-f1\\ m1a=m1gsin\theta +T-\mu \text{1}FN1\mathrm{......}\left(1\right)\\ \end{array}$

Applying Newton's second law to the first block, the net force along the vertical direction is,

  $\begin{array}{l}{\displaystyle \sum Fy}=m1g\text{ cos}\theta \\ 0=FN1\text{ -m1g cos}\theta \\ \end{array}$

Rewrite the equation for $FN1$ ,

  $\begin{array}{l}FN1\text{ = m1g cos}\theta \\ \end{array}$

Substitute $\text{m1g cos}\theta$ for $FN1$

  $m1a=m1gsin\theta +T-u1m1gcos\theta$

Thus, the tension in the string is,

  $T=m1a+u1m1gcos\theta -m1gsin\theta$

The frictional force acting on the second block is,

  $f2=\mu \text{2}FN2$

Here, $\mu \text{2}$ is coefficient of Kinetic friction between the second block and the incline and in

normal force on the second block.

Applying Newton's second law to the second block, the net force along the horizontal direction is,

  $$

Applying Newton's second law to the first block, the net force along the vertical direction is,

  $\begin{array}{l}{\displaystyle \sum FY}=FN2-m2gcos\theta \\ 0=FN2–m2gcos\theta \end{array}$

Rewrite the equation for F_N2.

  $FN2=m2gcos\theta$

Substitute $FN2=m2gcos\theta$

  $\begin{array}{l}\\ m2a=m2gsin\theta -T-\mu 2m2gcos\theta \end{array}$

Thus, the tension in the string is,

  $\begin{array}{l}\\ T=m2gsin\theta -\mu 2m2gcos\theta -m2a\end{array}$

Compare the equations of tensions, we get

  $\begin{array}{l}m1a-m1gsin\theta +\mu 1m1gcos\theta =m2gsin\theta -\mu \text{2}m2gcos\theta -m2a\\ m1a+m2a=m1gsin\theta +\mu 1m1gsin\theta -\mu 1m1gcos\theta -u2m2gcos\theta \\ \left(m1+m2\right)a=\left(m1+m2\right)gsin\theta -\left(\mu 1m1+\mu 2m2\right)gcos\theta \\ a=g\left[\sin \theta -\left(\frac{\mu 1m1+\mu 2m2}{m1+m2}\right)\cos \theta \right]\end{array}$

Thus, the magnitude of the common acceleration of the two blocks is as follows.

  $$

Substitute 9.8 m/s² for g, 1.0 for µ₁, 1.0 kg for m₁, 2.0 kg for m₂, 30.0° for θ, and 0.50 for µ₂,

  $\begin{array}{l}\\ \left|a\right|=\left(9.8m/s^2\right)\left[sin30.0°-\left(\frac{(\left(1.0\right)\left(1.0kg\right)+\left(0.5\right)\left(2.0kg\right)}{1.0kg+2.0kg}\right)cos30.0°\right]\\ =0.758m/s^2\end{array}$

Therefore, the acceleration of each block is 0.758 m/s²

Conclusion: By using Newton's second law acceleration can be determined.