Block Ain Fig. 6-56 has mass ${\mathit{m}}_{\mathbf{A}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$, and block Bhas mass ${\mathit{m}}_{\mathbf{B}\mathbf{}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$.The coefficient of kinetic friction between block B and the horizontal plane is ${\mathbf{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}$.The inclined plane is frictionless and at angle $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{°}$. The pulley serves only to change the direction of the cord connecting the blocks. The cord has negligible mass. Find

(a) the tension in the cord and

(b) the magnitude of the acceleration of the blocks.

• Mass of block A:$m_A=4.0\mathrm{kg}$ .
• Mass of block B:$m_B=2.0\mathrm{kg}$.
• The coefficient of kinetic friction between block B and the horizontal plane is:${\mu }_k=0.50$ .

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

Formula:

F = ma

Applying Newton’s second law:

For block A:

$$\begin{gathered} m_{A}g\sin \left({30}^a\right)-T=m_{A}a \\ -T=m_{A}g\sin \left(30°\right)-m_{A}a \end{gathered}$$

(i)

For block B:

$$\begin{gathered} T-f=m_{B}a \\ T=f+m_{B}a \end{gathered}$$

(ii)

(Where a is the acceleration of both blocks and f is the force due to friction)

Equating T from both equations:

$m_{A}g\sin (30°)-m_{A}a=f+m_{B}a$

The magnitude of the acceleration of the blocks can be calculated as:

Solving the above equation for acceleration a as:

$$\begin{gathered} m_{A}g\sin (30°)-f=\left(m_A+m_B\right)a \\ {m}_{A}g\sin (30°)-{\upsilon }_k\times F_{n2}=\left(m_A+m_B\right)a \\ a=\frac{m_{A}g\sin (30°)-{\upsilon }_k\times F_{n2}}{\left(m_A+m_B\right)} \\ =\frac{m_{A}g\sin (30°)-{\upsilon }_k\times \left(m_{B}g\right)}{\left(m_A+m_B\right)} \end{gathered}$$

Substituting the values in the above expression, and we get,

$$\begin{gathered} \mathrm{a}=\frac{\left(4.0\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \sin (30°)\right)-\left(0.50\times 2.0\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\right)}{4.0\mathrm{kg}+2.0\mathrm{g}} \\ \mathrm{a}=\frac{9.8\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{6.0\mathrm{kg}} \\ \mathrm{a}=1.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the magnitude of the acceleration is$\mathrm{a}=1.6\mathrm{m}/{\mathrm{s}}^2$ .

Using equation (ii), we get,

$$\begin{gathered} T=f+m_{B}a \\ =\left({\mu }_k\times F_{n2}\right)+m_{B}a \end{gathered}$$

Substituting the values in the above expression, and we get,

$$\begin{gathered} \mathrm{T}=\left(0.50\times 2.0\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\right)+\left(2.0\mathrm{kg}\times 1.6\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{T}=13\mathrm{N} \end{gathered}$$

Thus, the tension in the rope is 13 N.