In Fig. 6-46, a box of ant aunts (total mass m₁ = 1.65kg) and a box of ant uncles (total mass m = 3.30kg) slide down an inclined plane while attached by a massless rod parallel to the plane. The angle of incline is $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$. The coefficient of kinetic friction between the aunt box and the incline is ${\mathit{\mu }}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{226}$; that between the uncle box and the incline is ${\mathit{\mu }}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{113}$. Compute

(a) the tension in the rod and

(b) the magnitude of the common acceleration of the two boxes.

(c) How would the answers to (a) and (b) change if the uncles trailed the aunts?

• Mass of box of ant aunts, m₁ = 1.65kg.
• Mass of box of ant uncles, m₂ = 3.30kg.
• The angle of inclination,$\theta =30.0^{°}$.

The coefficient of kinetic friction between the aunt box and the incline is,${\mu }_1=0.226$The coefficient of kinetic friction between the uncle’s box and the incline is, ${\mu }_2=0.113$.

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object.

First, draw the free-body diagram of the two boxes. Then applying Newton's second law, solve the given problem.

T is the magnitude of the force in the rod (when T > 0 the rod is said to be in tension and when T < 0 the rod is under compression), ${\stackrel{\rightharpoonup }{F}}_{N2}$is the normal force on box 2 (the uncle box), ${\stackrel{\rightharpoonup }{F}}_{N1}$is the normal force on the aunt box (box 1), ${\stackrel{\rightharpoonup }{f}}_1$is kinetic friction force on the aunt box, and ${\stackrel{\rightharpoonup }{f}}_2$is the kinetic friction force on the uncle box.

Also, m₁ = 1.65kg is the mass of the aunt box and m₂ = 3.30kg is the mass of the uncle box (which is a lot of ants!).

For each block, we take +x downhill (which is toward the lower-right in these diagrams) and +y in the direction of the normal force.

Applying Newton's second law to the x and y directions of first box 2 and next box 1, we arrive at four equations:

$$\begin{gathered} m_{2}g\sin \theta -f_2-T=m_{2}a \\ {F}_{N2}-m_{2}g\cos \theta =0 \\ {F}_{N1}-m_{1}g\cos \theta =0 \end{gathered}$$

Which, when combined with equations $f_1={\mu }_1{F}_{N1}$where ${\mu }_1=0.226$and $f_2={\mu }_2{F}_{N2}$where ${\mu }_2=m_{1}g\cos \theta =0$, fully describe the dynamics of the system.

We solve the above equations for the tension and obtain,

$T=\left(\frac{m_2{m}_{1}g}{m_2+m_1}\right)\left({\mu }_1-{\mu }_2\right)\cos \theta$

Substitute the values, and we get,

$\begin{array}{rcl}T& =& \left(\frac{\left(3.30\right)\left(1.65\right)\left(9.8\right)}{3.30+1.65}\right)\left(0.226-0.113\right)\cos {30}^{°}\\ & =& 1.05N\end{array}$

Thus, the tension in the rod is 1.05N.

These equations lead to an acceleration equal to,

$a=g\left(\sin \theta -\left(\frac{{\mu }_2{m}_2+{\mu }_1{m}_1}{m_2+m_1}\right)\cos \theta \right)$

Substitute the values, and we get,

$\begin{array}{rcl}a& =& 9.8\left(\sin {30}^{°}-\left(\frac{0.113\times 3.30+0.226\times 1.65}{3.30+1.65}\right)\cos {30}^{°}\right)\\ & =& 3.62m/s^2\end{array}$

Thus, the magnitude of the common acceleration is 3.62m/s².

c)

Reversing the blocks is equivalent to switching the labels. We see from our algebraic result in part (a) that this gives a negative value for T (equal in magnitude to the result we got before).

Thus, the situation is as it was before, except that the rod is now in a state of compression.