Question

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed $\mathbf{1100}\mathbf{}\mathbf{N}$. The coefficient of static friction between the box and the floor is $\mathbf{0}\mathbf{.}\mathbf{35}$. (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

Short answer

(a) The angle between the cable and the horizontal in order to pull the greatest possible amount of sand is${}^{{\mathbf{19}}^{\mathbf{0}}}$

(b) The weight of the sand and box is${}^{\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{N}}$

Step-by-step solution

Given

Tension on the cable,${}^{\mathbf{T}\mathbf{=}\mathbf{1100}\mathbf{}\mathbf{N}}$

Coefficient of static friction between the table and box, ${}^{{\mathbf{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}}$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. First, draw the free body diagram of the box of sand and then apply Newton's 2nd law of motion to find the angle of the cable and weight of the sand box.

Formula:

$F_{net}=\sum ma$

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

Determining the free body diagram

Free Body Diagram of the sand box:

(a) Determining the angle between the cable and the horizontal

Initially the box is at rest so its acceleration along the vertical and horizontal direction is 0.By using Newton’s 2nd law of motion along the vertical direction,

$$\begin{gathered} \mathrm{N}+\mathrm{T}\sin \left(\mathrm{\theta }\right)-\mathrm{W}=0 \\ \mathrm{N}=\mathrm{mg}=\mathrm{T}\sin \left(\mathrm{\theta }\right) \end{gathered}$$

Along horizontal direction,

$$\begin{gathered} Tcos\left(\theta \right)=f_s \\ Tcos\left(\theta \right)={\mu }_{s}N \\ Tcos\left(\theta \right)={\mu }_s\left(mg-Tsin\left(\theta \right)\right) \\ \frac{T}{g}\left(\frac{cos\left(\theta \right)}{{\mu }_s}+sin\left(\theta \right)\right)=m \end{gathered}$$

It is given that sand should be greatest amount, by using the calculus method; take the change in mass (m) with respect to the angle $\left(\theta \right)$as 0,

$$\begin{gathered} \frac{dm}{d\theta }=\frac{T}{g}\left(\frac{-\sin \left(\theta \right)}{{\mu }_s}+\cos \left(\theta \right)\right) \\ =0 \\ \sin \left(\theta \right)={\mu }_s\cos \left(\theta \right) \\ \tan \left(\theta \right)={\mu }_s \\ \theta ={\tan }^{-1}\left({\mu }_s\right) \\ ={\tan }^{-1}\left(0.35\right) \\ =19.{29}^0 \\ \approx {19}^0 \end{gathered}$$

Hence, the angle between the cable and the horizontal in order to pull the greatest possible amount of sand is${19}^0$

(b) Determining the weight of the sand and box 

The mass of the box and sand is,

$$\begin{gathered} m=\frac{T}{g}\left(\frac{\cos \left(\theta \right)}{{\mu }_s}+\sin \left(\theta \right)\right) \\ m=\frac{1100}{9.81}\left(\frac{\cos \left(19\right)}{0.35}+\sin \left(19\right)\right) \\ =339.4kg \end{gathered}$$

And weight of this system,

$$\begin{gathered} w=mg \\ =(339.4)(9.81) \\ =3.33\times {10}^3\mathrm{N} \end{gathered}$$

Hence, the weight of the sand and box is${}^{3.33\times {10}^3\mathrm{N}}$