Question

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?

Short answer

(a) If Body A is at rest, then its acceleration is 0

(b) If Body A is moving up, it has acceleration ${}^{-3.9\mathrm{m}/{\mathrm{s}}^2}$

(c) If Body A is moving down, it has acceleration ${}^{-1.0\mathrm{m}/{\mathrm{s}}^2}$

Step-by-step solution

Given

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, ${}^{\mathbf{\theta }\mathbf{=}{\mathbf{40}}^{\mathbf{0}}}$

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. 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.

Determining the free body diagram

Free Body Diagram of the body A and body B:

(a) Determining the acceleration of A if A is initially at rest

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

(b) Determining the acceleration of A if A is initially at moving up the incline

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.

(c) Determining the moving down the incline

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,${}^{(-1.0\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{i}}}$