Chapter 10, Problem 101A

To determine

The force needed to push horizontally the can.

Answer to Problem 101A

The force needed to push horizontally the can is $82\text{ N}$ .

Explanation of Solution

Given:

The mass of the can is $m=24\text{ kg}$ .

The coefficient of kinetic friction between the can and the muddy grass is ${\mu }_{\text{k}}=0.27$ .

The coefficient of static friction is ${\mu }_{\text{s}}=0.35$ .

Formula used:

The expression for the friction force is,

$F_{\text{friction}}={\mu }_{\text{s}}{F}_{\text{N}}$

Here, ${\mu }_{\text{s}}$ is the static friction and $F_{\text{N}}$ is the normal force.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^2$ .

The normal force on the crate is,

$\begin{array}{l}{F}_N=mg\\ F_N=\left(\text{24 kg}\right)\times \left(9.8\text{m}/{\text{s}}^2\right)\\ F_N=235.2\text{ N}\end{array}$

The force needed to push horizontally to get the can to just start moving is

$\begin{array}{l}{F}_{\text{can}}=F_{\text{friction}}\\ F_{\text{can}}={\mu }_{\text{s}}{F}_{\text{N}}\\ F_{\text{can}}=0.35\times \left(235.2\text{ N}\right)\\ F_{\text{can}}=82\text{ N}\end{array}$

Conclusion:

Thus, the force needed to push horizontally the can is $82\text{ N}$ .