Chapter 8, Problem 1FP

F8–1. Determine the friction developed between the 50-kg crate and the ground if a) P=200 N, and b) P=400 N. The coefficients of static and kinetic friction between the crate and the ground are, μ_s = 0.3 and, μ_k = 0.2.

To determine

The frictional force (F).

Answer to Problem 1FP

(a) The frictional force (F) is $\underset{\_}{160\text{N}}$.

(b) The frictional force (F) is $\underset{\_}{146\text{N}}$.

Explanation of Solution

Given:

The mass of the crate (m) is 50 kg.

The co-efficient of static friction $\left({\mu }_s\right)$ is 0.3.

The co-efficient of kinetic friction $\left({\mu }_k\right)$ is 0.2.

Show the free body diagram of the crate as in Figure (1).

Using Figure (1),

Determine the normal reaction (N) using the equation of equilibrium.

Along the vertical direction:

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ N-W-P\times \left(\frac{3}{5}\right)=0\\ N-mg-P\times \left(\frac{3}{5}\right)=0\end{array}$ (I)

Here, the acceleration due to gravity is g.

Determine the frictional force using the equation of equilibrium.

Along the horizontal direction:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F-P\left(\frac{4}{5}\right)=0\end{array}$ (II)

Conclusion:

Substitute 50 kg for m, $9.81{\text{m/sec}}^{\text{2}}$ for g, and 200 N for P in Equation (I).

$\begin{array}{l}N-50\times 9.81-200\times \left(\frac{3}{5}\right)=0\\ N=610.5\text{N}\end{array}$

Substitute 200 N for P in Equation (II).

$\begin{array}{l}F-200\times \left(\frac{4}{5}\right)=0\\ F=160\text{N}\end{array}$

Check:

Determine the maximum frictional force.

$F_{\max }={\mu }_{s}N$

Substitute 0.3 for ${\mu }_s$ and 610.5 N for N.

$\begin{array}{c}{F}_{\max }=0.3\times 610.5\\ =183.15\text{N}\end{array}$

The maximum frictional force (183.15 N) is greater than the frictional force (160 N); therefore the block will not slip.

Therefore, the frictional force is $\underset{\_}{160\text{N}}$.

To determine

The frictional force (F).

Answer to Problem 1FP

The frictional force (F) is $\underset{\_}{146\text{N}}$.

Explanation of Solution

Given:

The mass of the crate (m) is 50 kg.

The co-efficient of static friction $\left({\mu }_s\right)$ is 0.3.

The co-efficient of kinetic friction $\left({\mu }_k\right)$ is 0.2.

Show the free body diagram of the crate as in Figure (2).

Using Figure (2),

Determine the normal reaction (N) using the equation of equilibrium.

Along the vertical direction:

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ N-W-P\times \left(\frac{3}{5}\right)=0\\ N-mg-P\times \left(\frac{3}{5}\right)=0\end{array}$ (III)

Determine the frictional force using the equation of equilibrium.

Along the horizontal direction:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F-P\left(\frac{4}{5}\right)=0\end{array}$ (IV)

Conclusion:

Substitute 50 kg for m, $9.81{\text{m/sec}}^{\text{2}}$ for g, and 400 N for P in Equation (III).

$\begin{array}{l}N-50\times 9.81-400\times \left(\frac{3}{5}\right)=0\\ N=\text{730}\text{.5}\text{N}\end{array}$

Substitute 400 N for P in Equation (IV).

$\begin{array}{l}F-400\times \left(\frac{4}{5}\right)=0\\ F=320\text{N}\end{array}$

Check:

Determine the maximum frictional force.

$F_{\max }={\mu }_{s}N$

Substitute 0.3 for ${\mu }_s$ and 730.5 N for N.

$\begin{array}{c}{F}_{\max }=0.3\times 730.5\\ =219.15\text{N}\end{array}$

The maximum frictional force (219.15 N) is less than the frictional force (320 N); therefore the block will slip. Hence, find the frictional force using the co-efficient of kinetic friction.

Find the frictional force using the relation.

$F={\mu }_{k}N$

Substitute 0.2 for ${\mu }_k$ and 730.5 N for N.

$\begin{array}{c}F=0.2\times 730.5\\ =146\text{N}\end{array}$

Thus, the frictional force is $\underset{\_}{146\text{N}}$.