Chapter 5, Problem 50PQ

FIGURE P5.49 Problems 49 and 50.

Suppose the system of blocks in Problem 49 is initially held motionless and, when released, begins to accelerate. a. If m₁ = 7.00 kg, m₂ = 2.00 kg, and the magnitude of the acceleration of the blocks is 0.134 m /s², find the magnitude of the kinetic frictional force between the second block and the ledge. b. What is the value of the coefficient of kinetic friction between the block and the ledge?

To determine

What is the kinetic friction force between the second block and ledge?

Answer to Problem 50PQ

The kinetic friction force between the second block and ledge is $\text{60}\text{.8 N}$.

Explanation of Solution

The free body diagram is given below.

Applying Newton’s laws.

    $\Sigma F_y=F_{g1}-F_T=m_{1}a$                                                                                (I)

Here, $F_y$ is the force, $F_T$ is the tension force, $m_1$ is the mass, $F_{g1}$ is the gravitational force on $m_1$ and $a$ is the acceleration.

    $\Sigma F_x=F_T-F_{g2,x}-F_k=m_{2}a$                                                                       (II)

Here, $F_{g2,x}$ is the gravitational force on $m_2$ and $F_k$ is the friction force.

Write the equation for gravitational force.

    $F_g=mg$                                                                                              (III)

Here, $g$ is the acceleration due to gravity.

Conclusion:

Using equation III find the gravitational force on each object.

    $\begin{array}{c}{F}_{g1}=m_{1}g\\ =\left(7.00\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\\ =68.7\text{ N}\end{array}$

    $\begin{array}{c}{F}_{g2}=m_{1}g\\ =\left(2.00\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\\ =19.6\text{ N}\end{array}$

From the free body diagram the find the value of $F_{g2,x}$.

    $\begin{array}{c}{F}_{g2,x}=F_{g1}\sin \left(20°\right)\\ =\left(19.6\text{ N}\right)\sin \left(20°\right)\\ =6.70\text{ N}\end{array}$

Substitute the above values and $\text{0}{\text{.134 m/s}}^2$ for $a$ in equation I to get tension force.

    $\begin{array}{c}68.7\text{ N}-F_T=\left(7.00\text{ kg}\right)\left(\text{0}{\text{.134 m/s}}^2\right)\\ F_T=68.7\text{ N}-\left(7.00\text{ kg}\right)\left(\text{0}{\text{.134 m/s}}^2\right)\\ =67.8\text{ N}\end{array}$

Substitute above values in equation II to get friction force.

    $\begin{array}{c}68.7\text{ N}-6.70\text{ N}-F_k=\left(2.00\text{ kg}\right)\left(\text{0}{\text{.134 m/s}}^2\right)\\ 68.7\text{ N}-6.70\text{ N}-F_k=0.268\text{ N}\\ F_k=60.8\text{ N}\end{array}$

Therefore, the kinetic friction force between the second block and ledge is $\text{60}\text{.8 N}$.

To determine

Find the coefficient of kinetic friction force between the block and ledge.

Answer to Problem 50PQ

The coefficient of kinetic friction force between the block and ledge is $3.30$.

Explanation of Solution

Applying Newton’s laws for y direction on block 2.

    $\Sigma F_y=F_N-F_{g2,y}=0$                                                                       (IV)

Here, $F_{g2,y}$ is the gravitational force on block 2 in y direction and $F_N$ is the normal force.

Write the equation for friction force.

    $F_k={\mu }_k{F}_N$                                                                                               (V)

Here, $F_k$ is the friction force, ${\mu }_k$ is the coefficient of kinetic friction and $F_N$ is the normal force.

Conclusion:

From the free body diagram the find the value of $F_{g2,y}$.

    $\begin{array}{c}{F}_{g2,y}=F_{g2}\cos \left(20°\right)\\ =\left(19.6\text{ N}\right)\cos \left(20°\right)\\ =18.4\text{ N}\end{array}$

Substitute $18.4\text{ N}$ for $F_{g2,y}$ in equation IV.

    $\begin{array}{c}{F}_N-18.4\text{ N}=0\\ F_N=18.4\text{ N}\end{array}$

Substitute $18.4\text{ N}$ for $F_N$ and $\text{60}\text{.8 N}$ for $F_K$ in equation V to find coefficient of friction.

     $\begin{array}{c}{\mu }_k=\frac{F_k}{F_N}\\ =\frac{\text{60}\text{.8 N}}{18.4\text{ N}}\\ =3.30\end{array}$

Therefore, the coefficient of kinetic friction force between the block and ledge is $3.30$.