Q77P - University Physics with Modern Physics

Question

A block with mass${\mathbf{m}}_{\mathbf{1}}$is placed on an inclined plane with slope angle and is connected to a hanging block with mass ${\mathbf{m}}_{\mathbf{2}}$by a cord passing over a small, frictionless pulley (Fig. P5.74). The coefficient of static friction is , and the coefficient of kinetic friction is${\mathbf{\mu }}_{\mathbf{k}}$ (a) Find the value of for which the block of mass ${\mathbf{m}}_{\mathbf{1}}$moves up the plane at constant speed once it is set in motion. (b) Find the value of for which the block of mass ${\mathbf{m}}_{\mathbf{2}}$moves down the plane at constant speed once it is set in motion. (c) For what range of values of ${\mathbf{m}}_{\mathbf{2}}$will the blocks remain at rest if they are released from rest?

       

                     

Step-by-Step Solution

Answer

1. The value of the ${\mathrm{m}}_2$ when the block moves up the plane is $m_1\left(\sin \alpha +{\mu }_k\cos \alpha \right)$.
2. The value of the ${\mathrm{m}}_2$ when the block moves down the plane is $m_1\left(\sin \alpha +{\mu }_k\cos \alpha \right)$.
3. The range of  values of ${\mathrm{m}}_2$is $m_1\left(\sin \alpha +{\mu }_k\cos \alpha \right)<m_2<m_1\left(\sin \alpha +{\mu }_k\cos \alpha \right)$ .

Step 1: Identification of the given data

The given data can be listed below as:

The mass of the first block is ${\mathrm{m}}_1$.

The mass of the first block is ${\mathrm{m}}_2$.

The value of the slope angle is $\mathrm{\alpha }$.

The static friction’s coefficient is ${\mathrm{\mu }}_{\mathrm{s}}$.

The kinetic friction’s coefficient is ${\mathrm{\mu }}_{\mathrm{k}}$.

Step 2: Significance of the tension in physics

The tension is a force required to pull a string to transmit the force in the axial direction. Tension is described as the pair of action and reaction forces.

Step 3: (a) Determination of the value of if the first block moves up the plane

The free body diagram of the block moving upward is drawn below:

               

The downward tension of the cord’s is ${\mathrm{m}}_2\mathrm{g}$ where ${\mathrm{m}}_2$ is the mass of the second block and g is the acceleration due to gravity. However, the tension should overcome the friction as well as the gravitational force’s component in order to move the block at a constant speed. As the mass ${\mathrm{m}}_1$ is moving at an inclined direction then the weight of the block has two components such as ${\mathrm{m}}_1\mathrm{gcos\alpha }$ and ${\mathrm{m}}_{1}g\sin \alpha$ respectively. The frictional force is ${\mu }_k{m}_1\cos \alpha$.

 

The summation of the forces in the x direction is zero.

The equation of the summation of the forces in the direction is expressed as:

 $\sum _{ }{\mathrm{F}}_{\mathrm{x}}=0$

   ${\mathrm{m}}_2\mathrm{g}={\mathrm{m}}_1\mathrm{gsin\alpha }+{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_1\mathrm{gcos\alpha }$

Here, $\sum _{ }{\mathrm{F}}_{\mathrm{x}}$ is the summation of the forces in the x direction, ${\mathrm{m}}_2$ is the mass of the second block, ${\mathrm{m}}_1$ is the mass of the first block, $\alpha$ is the angle between the first and the second masses and g is the acceleration due to gravity.

The above equation can be expressed as:

$$\begin{gathered} {\mathrm{m}}_2\sin ={\mathrm{m}}_1\mathrm{sin\alpha }+{\mathrm{\mu m}}_1\mathrm{cos\alpha } \\ ={\mathrm{m}}_1\left(\mathrm{sin\alpha }+{\mathrm{\mu }}_{\mathrm{k}}\mathrm{cos\alpha }\right) \end{gathered}$$

Thus, the value of the ${\mathrm{m}}_2$ when the block moves up the plane is ${\mathrm{m}}_1\left(\mathrm{sin\alpha }+{\mathrm{\mu }}_{\mathrm{k}}\mathrm{cos\alpha }\right)$

Step 4: (b) Determination of the value of m 2 if the first block moves down the plane

The free body diagram of the block moving downward has been drawn below:

                 

The downward tension of the cord’s is ${\mathrm{m}}_2\mathrm{g}$ where ${\mathrm{m}}_2$ is the mass of the second block and g is the acceleration due to gravity. However, the tension should overcome the friction as well as the gravitational force’s component in order to move the block at a constant speed. As the mass is moving at an inclined direction then the weight of the block has two components such as and respectively. The frictional force is ${\mu }_k{m}_{1}g\cos \alpha$.

The summation of the forces in the direction is zero.

The equation of the summation of the forces in the direction is expressed as:

 $$\begin{gathered} \sum _{ }{\mathrm{F}}_{\mathrm{x}}=0 \\ {\mathrm{m}}_2\mathrm{g}={\mathrm{m}}_1\mathrm{gsin\alpha }+{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_1\mathrm{gcos\alpha } \end{gathered}$$

Here, $\sum _{ }{\mathrm{F}}_{\mathrm{x}}$  is the summation of the forces in the x direction,${\mathrm{m}}_2$ is the mass of the second block, ${\mathrm{m}}_1$ is the mass of the first block, $\alpha$ is the angle between the first and the second masses and g is the acceleration due to gravity.

The above equation can be expressed as:

$$\begin{gathered} {\mathrm{m}}_2={\mathrm{m}}_1\mathrm{sin\alpha }+{\mathrm{\mu }}_{\mathrm{k}}\mathrm{cos\alpha } \\ ={\mathrm{m}}_1\left(\mathrm{sin\alpha }+{\mathrm{\mu }}_{\mathrm{k}}\mathrm{cos\alpha }\right) \end{gathered}$$

Thus, the value of the ${\mathrm{m}}_2$ when the block moves down the plane is ${\mathrm{m}}_1\left(\mathrm{sin\alpha }+{\mathrm{\mu }}_{\mathrm{k}}\mathrm{cos\alpha }\right)$.       

Step 5: (c) Determination of the range of values of m 2

According to the above answers, when the block moves upward, it has the largest value. Moreover, when the block moves downward, then it has lowest value. Hence, the range of should lie between these two values.

Thus, the range of values ${\mathrm{m}}_2 \mathrm{is} {\mathrm{m}}_1\left(\mathrm{sin\alpha }-{\mathrm{\mu }}_{\mathrm{k}} \mathrm{cos\alpha }\right)<{\mathrm{m}}_2<{\mathrm{m}}_1\left(\mathrm{sin\alpha }-{\mathrm{\mu }}_{\mathrm{k}} \mathrm{cos\alpha }\right)$