Chapter 12, Problem 5IP

A football player pushes a 670-N blocking sled. The coefficient of static friction between sled and grass is 0.73, and the coefficient of kinetic friction between sled and grass is 0.68.

a. How much force must the player exert to start the sled in motion?

b. How much force is required to keep the sled in motion?

c. Answer the same two questions with a 100-kg coach standing on the back of the sled.

Summary Introduction

To determine: The force required to cause the sled in motion.

Answer to Problem 5IP

The force required to cause the sled in motion is greater than $489.1\text{N}$.

Explanation of Solution

Calculation:

Write an expression to find the force required to cause the sled in motion.

$F_s={\mu }_{s}R$

Here, $F_s$ is the force required to cause the sled in motion, ${\mu }_s$ coefficient of static friction, and $R$ is the normal reaction force.

Substitute 0.73 for ${\mu }_s$ and $670\text{N}$ for $R$ to find the force required to cause the sled in motion.

$\begin{array}{c}{F}_s=\left(0.73\right)\left(670\text{N}\right)\\ =489.1\text{N}\end{array}$

Therefore, the force required to cause the sled in motion is greater than $489.1\text{N}$.

Conclusion

Therefore, the force required to cause the sled in motion is greater than $489.1\text{N}$.

Summary Introduction

To determine: The force required to keep the sled in motion.

Answer to Problem 5IP

The force required to keep the sled in motion is greater than $455.6\text{N}$.

Explanation of Solution

Calculation:

Write an expression to find the force required to keep the sled in motion.

$F_k={\mu }_{k}R$

Here, $F_k$ is the force required to keep the sled in motion, ${\mu }_k$ coefficient of kinetic friction, and $R$ is the normal reaction force.

Substitute 0.68 for ${\mu }_k$ and $670\text{N}$ for $R$ to find the force required to keep the sled in motion.

$\begin{array}{c}{F}_k=\left(0.68\right)\left(670\text{N}\right)\\ =455.6\text{N}\end{array}$

Therefore, the force required to keep the sled in motion is greater than $455.6\text{N}$.

Conclusion

Therefore, the force required to keep the sled in motion is greater than $455.6\text{N}$.

Summary Introduction

To determine: The force required to cause the sled and coach in motion.

Answer to Problem 5IP

The force required to cause the sled and coach in motion is greater than $1205.2\text{N}$.

Explanation of Solution

Calculation:

Calculate the total reaction force.

$R_T=\text{Noraml}\text{reaction}\text{force}+\left(\text{mass}\text{of}\text{the}\text{coach}\times \text{acceleration}\text{due}\text{to}\text{gravity}\right)$

Here, $R_T$ is the total reaction force.

Substitute $670\text{N}$ for normal reaction force, 100 kg for mass of the coach, and $9.81\text{m}/{\text{s}}^{\text{2}}$ for acceleration due to gravity to find the total reaction force.

$\begin{array}{c}{R}_T=\text{670}\text{N}+\left(\text{100}\text{kg}\times \text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}\right)\\ =1651\text{N}\end{array}$

Therefore the total reaction force is $1651\text{N}$.

Write an expression to find the force required to cause the sled and coach in motion.

$F_s={\mu }_s{R}_T$

Here, $F_s$ is the force required to cause the sled and coach in motion and ${\mu }_s$ coefficient of static friction.

Substitute 0.73 for ${\mu }_s$ and $1651\text{N}$ for $R_T$ to find the force required to cause the sled and coach in motion.

$\begin{array}{c}{F}_s=\left(0.73\right)\left(1651\text{N}\right)\\ =1205.23\text{N}\\ \approx 1205.2\text{N}\end{array}$

Therefore, the force required to cause the sled and coach in motion is greater than $1205.2\text{N}$.

Conclusion

Therefore, the force required to cause the sled and coach in motion is greater than $1205.2\text{N}$.

Summary Introduction

To determine: The force required to keep the sled and coach in motion.

Answer to Problem 5IP

The force required to keep the sled and coach in motion is greater than $1122.7\text{N}$.

Explanation of Solution

Calculation:

Calculate the total reaction force.

$R_T=\text{Noraml}\text{reaction}\text{force}+\left(\text{mass}\text{of}\text{the}\text{coach}\times \text{acceleration}\text{due}\text{to}\text{gravity}\right)$

Here, $R_T$ is the total reaction force.

Substitute $670\text{N}$ for normal reaction force, 100 kg for mass of the coach, and $9.81\text{m}/{\text{s}}^{\text{2}}$ for acceleration due to gravity to find the total reaction force.

$\begin{array}{c}{R}_T=\text{670}\text{N}+\left(\text{100}\text{kg}\times \text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}\right)\\ =1651\text{N}\end{array}$

Therefore the total reaction force is $1651\text{N}$.

Write an expression to find the force required to keep the sled and coach in motion.

$F_k={\mu }_k{R}_T$

Here, $F_k$ is the force required to keep the sled and coach in motion and ${\mu }_k$ coefficient of kinetic friction.

Substitute 0.68 for ${\mu }_k$ and $1651\text{N}$ for $R_T$ to find the force required to keep the sled and coach in motion.

$\begin{array}{c}{F}_k=\left(0.68\right)\left(1651\text{N}\right)\\ =1122.68\text{N}\\ \approx 1122.7\text{N}\end{array}$

Therefore, the force required to keep the sled and coach in motion is greater than $1122.7\text{N}$.

Conclusion

Therefore, the force required to keep the sled and coach in motion is greater than $1122.7\text{N}$.