Chapter 2, Problem 74P

(a)

To determine

Whether the static friction can hold the safe in the place without sliding back down, if the movers stop pushing on the safe in Example 2.14.

Answer to Problem 74P

The static friction does not hold the safe in the place without sliding back down since the value of static frictional force is less than the net downward force.

Explanation of Solution

Figure 1 represents $x$ and $y$ components of the weight $W$.

Figure 2 represents the free body diagram of the safe in which the forces are represented by their respective x and y components with sign.

Write the expression for component of force of gravity acting on the safe.

    $W_{\text{x}}=mg\sin \theta$        (I)

Here, $W_{\text{x}}$ is the component of force of gravity acting on the safe, $m$ is the mass, and $g$ is the acceleration due to gravity.

Even after the mover stops pushing the block the safe still moves upwards, this motion has kinetic frictional force involved.

Write the expression for kinetic frictional force when the block moves upwards even after stop pushing the block.

    $f_{\text{k}}={\mu }_{\text{k}}mg\cos \theta$        (II)

Here, $f_{\text{k}}$ is the kinetic frictional force, ${\mu }_{\text{k}}$ is the coefficient of kinetic friction, $m$ is the mass, and $g$ is the acceleration due to gravity.

Write the expression for net downward force acting on the ramp.

    $f_{\text{net}}=W_{\text{x}}+f_{\text{k}}$        (III)

Here, $f_{\text{net}}$ is the net downward force acting on the ramp.

Write the expression for static frictional force acting on the safe.

    $f_{\text{s}}={\mu }_{\text{s}}mg\text{cos}\theta$

Here, $f_{\text{s}}$ is the static frictional force, and ${\mu }_{\text{s}}$ is the coefficient of static friction.

Conclusion:

Substitute $510\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, and $15°$ for $\theta$ in equation (I), to find $W_{\text{x}}$.

    $\begin{array}{c}{W}_{\text{x}}=\left(510\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\sin \left(15°\right)\\ =1294\text{N}\end{array}$

Substitute $510\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, $0.33$ for ${\mu }_{\text{k}}$ and $15°$ for $\theta$ in equation (II), to find $f_{\text{k}}$.

    $\begin{array}{c}{f}_{\text{k}}=\left(0.33\right)\left(510\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\cos \left(15°\right)\\ =1593\text{N}\end{array}$

Substitute $1294\text{N}$ for $W_{\text{x}}$, and $1593\text{N}$ for $f_{\text{k}}$ in equation (III), to find $f_{\text{net}}$.

    $\begin{array}{c}{f}_{\text{net}}=1294\text{N}+1593\text{N}\\ \text{=2887}\text{N}\end{array}$

Substitute $510\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, $0.42$ for ${\mu }_{\text{s}}$ and $15°$ for $\theta$ in equation (IV), to find $f_{\text{s}}$.

    $\begin{array}{c}{f}_{\text{s}}=\left(0.42\right)\left(510\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\text{cos}15°\\ =2027\text{N}\end{array}$

Therefore, the static friction does not hold the safe in the place without sliding back down since the value of static frictional force is less than the net downward force.

(b)

To determine

The minimum force required to hold the safe in place.

Answer to Problem 74P

The minimum force required to hold the safe in place is $\underset{\_}{860\text{N}}$.

Explanation of Solution

Write the expression for minimum force required to hold the safe in place.

    $f_{\text{min}}=f_{\text{net}}-f_{\text{s}}$        (V)

Here, $f_{\text{min}}$ is the minimum force required to hold the safe in place.

Conclusion:

Substitute $2887\text{N}$ for $f_{\text{net}}$, and $2027\text{N}$ for $f_{\text{s}}$ in equation (V), to find $f_{\text{min}}$.

    $\begin{array}{c}{f}_{\text{min}}=2887\text{N}-2027\text{N}\\ \text{=860}\text{N}\end{array}$

Therefore, the minimum force required to hold the safe in place is $\underset{\_}{860\text{N}}$.