Question

Question: A crate of weight ${\mathit{F}}_{\mathbf{g}}$is pushed by a force $\overrightarrow{\mathbf{P}}$on a horizontal floor as shown in Figure P5.89. The coefficient of static friction is $\mathit{\mu }$, and $\overrightarrow{\mathit{P}}$is directed at angle $\mathit{\theta }$below the horizontal. (a) Show that the minimum value of $\mathit{P}$ that will move the crate is given by

$\mathit{P}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{g}}{\mathbf{F}}_{\mathbf{g}}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{\theta }}{\mathbf{1}\mathbf{-}{\mathbf{\mu }}_{\mathbf{g}}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta }}$

(b) Find the condition on $\mathit{\theta }$in terms of ${\mathit{\mu }}_{\mathbf{s}}$for which motion of the crate is impossible for any value of $\mathit{P}$.

Short answer

(a) The required expression $\mathrm{P}\le \frac{{\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{g}}\mathrm{sec\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\mathrm{tan\theta }}$is obtained for the minimum force required to move crate.

(b) The condition $\tan \theta <\frac{1}{{\mu }_s}$is not achieved, therefore any value of $P$can’t move the crate.

Step-by-step solution

Writing the given data from the question

Consider the given data as below.

Weight of the crate is $F_g$.

The force $P$ is used to push a crate on horizontal floor.

The force is directed at angle $\theta$.

Determining the formula

To calculate the minimum force to move the crate and condition on $\mathit{\theta }$in terms of ${\mathit{\mu }}_{\mathbf{s}}$use the expression for the static friction as follow.

${\mathit{f}}_{\mathbf{s}}\mathbf{=}{\mathit{\mu }}_{\mathbf{s}}\mathit{n}$

Here, $\mathit{n}$is the magnitude of the normal force.

(a) Calculating the expression for the minimum force to move the crate:

Consider the free body diagram as shown below.

Consider the equilibrium condition in horizontal direction.

$$\begin{gathered} \sum _{}^{}{F}_x=0 \\ P\cos \theta -f_s=0 \\ P\cos \theta =f_s \end{gathered}$$

Consider the equilibrium condition in vertical direction.

$\begin{array}{ll}\sum _n{F}_y=0& \\ n-P\sin \theta -F_g=0& \\ n=P\sin \theta +F_g& \end{array}$

The static friction between the object and surface is given by

$f\le {\mu }_{s}n$

Substitute $P\cos \theta$for $f_s$and $P\sin \theta +F_g$for $n$into the above equation.

$$\begin{gathered} P\cos \theta \le {\mu }_s\left(P\sin \theta +F_g\right) \\ P\cos \theta \le {\mu }_{s}P\sin \theta +{\mu }_s{F}_g \\ P\cos \theta -{\mu }_{s}P\sin \theta \le {\mu }_s{F}_g \\ P\left(\cos \theta -{\mu }_s\sin \theta \right)\le {\mu }_s{F}_g \end{gathered}$$

Divide the above equation by $\cos \theta$.

$$\begin{gathered} P\left(\frac{\cos \theta -{\mu }_s\sin \theta }{\cos \theta }\right)\le \frac{{\mu }_s{F}_g}{\cos \theta } \\ P\left(1-\frac{{\mu }_s\sin \theta }{\cos \theta }\right)\le \frac{{\mu }_s{F}_g}{\cos \theta } \\ P\left(1-{\mu }_s\tan \theta \right)\le {\mu }_s{F}_{g}sec\theta \\ P\le \frac{{\mu }_s{F}_{g}sec\theta }{\left(1-{\mu }_s\tan \theta \right)} \end{gathered}$$

Hence, the required expression $P\le \frac{{\mu }_s{F}_{g}sec\theta }{1-{\mu }_s\tan \theta }$ is obtained for the minimum force required to move crate.

 Step 4: (b) Determining the condition on θ in terms of μs

Define the condition on in terms of for which motion of the crate is impossible for any value of .

To set the crate into the motion, the horizontal component must overcome the frictional force.

$f_s\ge {\mu }_{s}n$

Substitute $P\cos \theta$for $f_s$and $P\sin \theta +F_g$for $n$into the above equation.

$$\begin{gathered} P\cos \theta \ge {\mu }_s\left(P\sin \theta +F_g\right) \\ P\cos \theta \ge {\mu }_{s}P\sin \theta +{\mu }_s{F}_g \\ P\cos \theta -{\mu }_{s}P\sin \theta \ge {\mu }_s{F}_g \\ P\left(\cos \theta -{\mu }_s\sin \theta \right)\ge {\mu }_s{F}_g \end{gathered}$$

The condition is satisfied only when $\cos \theta -{\mu }_s\sin \theta >0$.

$$\begin{gathered} \cos \theta -{\mu }_s\sin \theta >0 \\ \cos \theta >{\mu }_s\sin \theta \\ \tan \theta <\frac{1}{{\mu }_s} \end{gathered}$$

Since the condition $\tan \theta <\frac{1}{{\mu }_s}$is not achieved, therefore any value of $P$can’t move the crate.