Q 71P - University Physics with Modern Physics

Question

Question: Two boxes connected by a light horizontal rope are on a horizontal surface (Fig. E5.37). The coefficient of kinetic friction between each box and the surface is ${\mathit{\mu }}_{\mathbf{K}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{30}$. Box has mass 5.00kg, and box A has mass m. A force F with magnitude 40.0N and direction $\mathbf{53}\mathbf{.}\mathbf{1}\mathbf{°}$ above the horizontal is applied to the 5.00kg box, and both boxes move to the right with $\mathbf{a}=\mathbf{1}.\mathbf{50} \mathbf{m}/{\mathbf{s}}^{\mathbf{2}}$. (a) What is the tension T in the rope that connects the boxes? (b) What is m?

Step-by-Step Solution

Answer

a) The tension in the rope is 11.4N.

b) The value of m is 2.57kg.

Step 1: Identification of the given data

The given data can be listed below as:

The frictional coefficient amongst the surface and the box is ${\mu }_K=0.30$

The mass of the box B is $m_1=5.00 \text{kg}$

The mass of the box A is m

The magnitude of the force is $F=40.0 \text{N}$$F=40.0 \text{N}$

The force is directed at an angle of $\theta =53.1°$

The acceleration of the box is $a=1.50 {\text{m/s}}^{\text{2}}$

Step 2: Significance of the friction force

The friction force mainly opposes the motion that is relative between two objects which are moving. The friction force mainly transfers one energy to another energy.

Step 3: (a) Determination of the tension in the rope

The free body diagram of the box B is drawn below,

Here, the box B exerts a force $m_{1}g$ in the downward direction. The normal force exerted by the box is $n_B$ and the force on the box is F. The force has two components such as $F\cos \theta$ and $F\sin \theta$ respectively and the value of the angle is $53.1°$. The tension exerted by the rope is T and the frictional force is $f_k$.

According to the diagram, the summation of all the forces acting in the x and the y direction is zero.

The equation of the summation of all the forces acting in the x direction is expressed as:

$$\begin{gathered} \sum F_x=0 \\ F\cos \theta -T-f_k-m_{1}a=0 \\ T=F\cos \theta -f_k-m_{1}a \end{gathered}$$

Here, $\sum F_x$ is the summation of all the forces acting in the x direction, F is the magnitude of the force, $\theta$ is the angle of the direction of the force, T is the tension, $f_k$ is the friction of the box, $m_1$ is the mass of the box B and a is the acceleration of the box.

 

The equation of the summation of all the forces acting in the y direction is expressed as:

$$\begin{gathered} \sum F_y=0 \\ F\sin \theta +n_B-m_{1}g=0 \\ {n}_B=m_{1}g-F\sin \theta \end{gathered}$$

Here, $\sum F_y$ is the summation of all the forces acting in the y direction, $n_B$  is the normal reaction force on box B and g is the acceleration due to gravity.

 

The equation of the friction of the box is expressed as:

 $f_k={\mu }_k{n}_B$

Substitute $m_{1}g-F\sin \theta$ for $n_B$ in the equation of friction.

$f_k={\mu }_k\left(m_{1}g-F\sin \theta \right)$ 

Substitute ${\mu }_k\left(m_{1}g-F\sin \theta \right)$ for $f_k$ in the equation of tension.

$\begin{array}{rcl}T& =& F\cos \theta -{\mu }_k\left(m_{1}g-F\sin \theta \right)-m_{1}a\\ & =& F\cos \theta -{\mu }_k{m}_{1}g-{\mu }_{k}F\sin \theta -m_{1}a\end{array}$

Substitute 40.0N for F, $53.1°$ for $\theta$, 0.30 for ${\mu }_k$, 5.00kg for $m_1$, $9.8 {\text{m/s}}^{\text{2}}$ for g and $1.50 {\text{m/s}}^{\text{2}}$ for a in the above equation.    

$\begin{array}{rcl}T& =& \left(40.0 \text{N}\right)\cos 53.1°-\left(0.30\right)\left(5.00 \text{kg}\right)\left(9.8 {\text{m/s}}^{\text{2}}\right)-\left(0.30\right)\left(40.0 \text{N}\right)\sin 53.1°\\ & & -\left(5.00 \text{kg}\right)\left(1.50 {\text{m/s}}^{\text{2}}\right)\\ & =& 24.01 \text{N} -14.7 \text{kg}·{\text{m/s}}^{\text{2}}-9.59 \text{N} -\text{7.5} \text{kg}·{\text{m/s}}^{\text{2}}\\ & =& 24.01 \text{N}-14.7 \text{kg}·{\text{m/s}}^{\text{2}}\times \left(\frac{\text{1} \text{N}}{\text{1} \text{kg}·{\text{m/s}}^{\text{2}}}\right)-9.59 \text{N}-7.5 \text{kg}·{\text{m/s}}^{\text{2}}\times \left(\frac{\text{1} \text{N}}{\text{1} \text{kg}·{\text{m/s}}^{\text{2}}}\right)\\ & =& 11.4 \text{N}\end{array}$

Thus, the tension in the rope is 11.4N.

Step 4: (b) Determination of the value of m.

The free body diagram of the box A is drawn below,

In the above diagram, the box A exerts a force mg in the ground. The normal force on the box A is $n_A$. The tension of the box is T and the frictional force is $f_k$.

According to the diagram, the summation of all the forces acting in the x and the y direction is zero.

The equation of the summation of all the forces acting in the x direction is expressed as:

$$\begin{gathered} \sum F_x=0 \\ T-f_k=ma \end{gathered}$$

$\therefore m=\frac{T-f_{kB}}{a}.......\left(1\right)$

Here, $\sum F_x$ is the summation of all the forces acting in the x direction, F is the magnitude of the force, $\theta$ is the angle of the direction of the force, T is the tension, $f_{kB}$ is the friction of the box, m is the mass of the box A and a is the acceleration of the box.

 

The equation of the summation of all the forces acting in the y direction is expressed as:

$$\begin{gathered} \sum F_y=0 \\ {n}_A-mg=0 \\ {n}_A=mg \end{gathered}$$

Here, $\sum F_y$ is the summation of all the forces acting in the y direction, $n_A$ is the normal reaction force on box B and g is the acceleration due to gravity.

The equation of the friction of the box B is expressed as:

$f_{kB}={\mu }_k{n}_A$

Substitute mg for $n_A$ in the equation of friction.

$f_{kB}={\mu }_{k}mg$

Substitute ${\mu }_{k}mg$ for $f_{k B}$ in equation (1)

$$\begin{gathered} m=\frac{T-{\mu }_{k}mg}{a} \\ ma=T-{\mu }_{k}mg \\ m\left(a+{\mu }_{k}g\right)=T \\ m=\frac{T}{\left(a+{\mu }_{k}g\right)} \end{gathered}$$

Substitute 11.4N for T, $1.50 {\text{m/s}}^{\text{2}}$ for a, 0.30 for ${\mu }_k$ and $9.8 {\text{m/s}}^{\text{2}}$ for g in the above equation. 

$\begin{array}{rcl}m& =& \frac{\text{11.4} \text{N}}{\left(\text{1.50} {\text{m/s}}^{\text{2}}\text{ + }\left(\text{0.30}\right)\left(\text{9.8} {\text{m/s}}^{\text{2}}\right)\right)}\\ & =& \frac{\text{11.4} \text{N}}{\left(\text{1.50} {\text{m/s}}^{\text{2}}\text{ + }\left(2.94 {\text{m/s}}^{\text{2}}\right)\right)}\\ & =& \frac{\text{11.4} \text{N}}{\left(4.44 {\text{m/s}}^{\text{2}}\right)}\\ & =& 2.57 {\text{Ns}}^{\text{2}}\text{/m}\times \frac{\text{1} \text{kg}·{\text{m/s}}^{\text{2}}}{\text{1} \text{N}}\\ & =& 2.57 \text{kg}\end{array}$

Thus, the value of m is 2.57kg.