Body A in Fig. 6-33 weighs $\mathbf{102}\mathbf{}\mathit{N}$, and body B weighs $\mathbf{32}\mathbf{}\mathbf{N}$. The coefficients of friction between A and the incline are ${\mathbf{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{56}$and ${\mathbf{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}$. Angle $\mathbf{\theta }$is ${\mathbf{40}}^{\mathbf{0}}$. Let the positive direction of an $\mathit{x}$axis be up the incline. In unit-vector notation, what is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

Weight of Body A,${}^{\mathbf{W}\mathbf{=}\mathbf{102}\mathbf{}\mathbf{N}}$

Weight of Body B,${}^{\mathbf{W}\mathbf{}\mathbf{\text{'}}\mathbf{=}\mathbf{32}\mathbf{}\mathbf{N}}$

Coefficient of static friction between the incline and Body A,${}^{{\mathbf{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{56}}$

Coefficient of kinetic friction between the incline and Body A, ${}^{{\mathbf{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}}$

Angle of incline, ${}^{\mathbf{\theta }\mathbf{=}{\mathbf{40}}^{\mathbf{0}}}$

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. To find the acceleration of the body A, use Newton’s 2nd law of motion.

Formula:

$F_{net}=\sum ma$

where,Fis the force,mis mass andais an acceleration.

Free Body Diagram of the body A and body B:

If Body A is at rest, then its acceleration is 0.

By using Newton’s 2nd law of motion along the vertical direction (along y) to the system,(positive x axis along the incline and the positive y along the vertical direction),

$$\begin{gathered} N-W\cos \left(40\right)=0 \\ N=W\cos \left(40\right) \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} f_k={\mu }_{k}N \\ ={\mu }_{k}W\cos \left(40\right) \end{gathered}$$

Now, applying Newton’s 2nd along horizontal direction to the Body A having mass m,

$$\begin{gathered} \mathrm{T}-{\mathrm{f}}_{\mathrm{k}}-\mathrm{Wsin}\left(40\right)=\mathrm{ma} \\ \mathrm{T}-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{Wcos}\left(40\right)-\mathrm{Wsin}\left(40\right)=\mathrm{ma} \end{gathered}$$ (i)

Similarly apply to the Body B having mass m’,

$W\text{'}-T=m\text{'}a\left(ii\right)$

Adding equation (i) and (ii)

$$\begin{gathered} W^{\text{'}}-{\mu }_{k}W\cos \left(40\right)-W\sin \left(40\right)=(m+m^{\text{'}})a \\ a=\frac{W^{\text{'}}-{\mu }_{k}W\cos \left(40\right)-W\sin \left(40\right)}{(m+m^{\text{'}})} \\ =\frac{32-(0.25)\left(102\right)\cos \left(40\right)-\left(102\right)\sin \left(40\right)}{\left({\displaystyle \frac{102}{9.8}}+{\displaystyle \frac{32}{9.8}}\right)} \\ =-3.9m/s^2 \end{gathered}$$

So, if Body A is moving up it has acceleration ${}^{(-3.9\mathrm{m}/{\mathrm{s}}^2})\hat{\mathrm{i}}$, negative sign says that it is slowing down during the motion.

By using Newton’s 2nd law of motion along the vertical direction (along y) to the system,(positive x axis along the incline and the positive y along the vertical direction),

$$\begin{gathered} N-W\cos \left(40\right)=0 \\ N=W\cos \left(40\right) \end{gathered}$$

Thus, the frictional force is,

$$\begin{gathered} f_k={\mu }_{k}N \\ ={\mu }_{k}W\cos \left(40\right) \end{gathered}$$

Now, applying Newton’s 2nd along horizontal direction to the Body A having mass m,

$$\begin{gathered} T+f_k-W\sin \left(40\right)=ma \\ T+{\mu }_{k}W\cos \left(40\right)-W\sin \left(40\right)=ma\left(iii\right) \end{gathered}$$

Similarly apply to the Body B having mass m’,

$W\text{'}-T=m\text{'}a\left(iv\right)$

Adding equation (iii) and (iv),

$$\begin{gathered} Wsin\left(40\right)-{\mu }_{k}Wcos\left(40\right)+W^{\text{'}}=(m+m^{\text{'}})a \\ a=\frac{{\mu }_{k}wcos\left(40\right)-Wsin\left(40\right)+W^{\text{'}}}{(m+m^{\text{'}})} \\ =\frac{(0.25)\left(102\right)cos\left(40\right)-\left(102\right)sin\left(40\right)+32}{\left({\displaystyle \frac{102}{9.8}}+{\displaystyle \frac{32}{9.8}}\right)} \\ =-1.0m/s^2 \end{gathered}$$

The acceleration is again downhill the plane, that is,${}^{(-1.0\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{i}}}$