Question

A block is pushed across a floor by a constant force that is applied at downward angle ${}^{\mathbf{\theta }}$(Fig. 6-19). Figure 6-36 gives the acceleration magnitude a versus a range of values for the coefficient of kinetic friction ${{}^{\mathbf{\mu }}}_{\mathbf{k}}$between block and floor: ${}^{{\mathbf{a}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$, ${}^{{\mathbf{\mu }}_{\mathbf{k}\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}}$, ${}^{{\mathbf{\mu }}_{\mathbf{k}\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}}$.What is the value of ${}^{\mathbf{\theta }}$?

Short answer

The magnitude of an angle is${}^{60°}$

Step-by-step solution

Given

From the graph,

$$\begin{gathered} {\mu }_{k_1}=0at{a}_1=3.0\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{\mu }}_{{\mathrm{k}}_1}=0.20\mathrm{at}{\mathrm{a}}_2=0\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{\mu }}_{{\mathrm{k}}_1}=0.40\mathrm{at}{\mathrm{a}}_3=-3.0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Determining the concept

To find the angle ${}^{\mathbf{\theta }}$ use Newton's 2nd law of motion. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\underset{}{\overset{}{\sum ma}}$

where, F is the net force, m is mass and a is an acceleration.

Determining the free body diagram

Free Body Diagram of the Block:

Determining the magnitude of the angle θ

By using the Newton’s 2nd law along the vertical direction to the block A,(positive x axis along the right and the positive y along the vertical direction),

$$\begin{gathered} F_N-F\sin \theta -mg=0 \\ {F}_N=F\sin \theta +mg \end{gathered}$$

Thus, the kinetic frictional force,

$$\begin{gathered} f_k={\mu }_k{F}_N \\ ={\mu }_k\left(F\sin \theta +mg\right) \end{gathered}$$

Similarly, to the horizontal direct

$$\begin{gathered} F\cos \theta -f_k=ma \\ F\cos \theta -{\mu }_k\left(F\sin \theta +mg\right)=ma \\ a\frac{F}{m}\left(\cos \theta -{\mu }_k\sin \theta \right)-{\mu }_{k}g \\ \mathrm{At},{\mu }_k=0and{a}_1=3.0\mathrm{m}/{\mathrm{s}}^2, \\ 3.0=\frac{F}{m}\left(cos\theta \right) \\ \mathrm{At},{\mu }_{k_2}=0.20and{a}_2=0\mathrm{m}/{\mathrm{s}}^2 \\ 0=\frac{F}{m}\left(cos\theta -\left(0.20\right)\sin \theta \right)-\left(0.20\right)g \\ 0=\frac{F}{m}\left(\cos \theta \right)-\left(020\right)\frac{F}{m}\sin \theta -\left(0.20\right)g \\ 0=3.0-\left(0.20\right)\frac{F}{m}\sin \theta -\left(0.20\right)g \end{gathered}$$ (i)

From equation (i),

$$\begin{gathered} \left(0.20\right)\frac{F}{m}\sin \theta =3.0-\left(0.20\right)g \\ \left(0.20\right)\frac{\left(3.0\right)}{\left(\cos \theta \right)}\sin \theta =3.0-\left(0.20\right)g \\ \left(0.6\right)\tan \theta =1.04 \\ \tan \theta =\frac{1.04}{0.6} \\ =1.73 \\ \theta =60° \end{gathered}$$

Hence, the magnitude of an angle is$60°$