Two blocks, of weights 3.6 Nand 7.2 N, are connectedby a massless string and slide down a$\mathbf{30}\mathbf{°}$inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10, and the coefficient between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the taut string.

Consider the two blocks as A and B:

Weight of block A,${\mathrm{W}}_{\mathrm{A}}=3.6\mathrm{N}$

Mass of block A,$M_A=0.37\mathrm{kg}$

Weight of block B,${\mathrm{W}}_B=7.2\mathrm{N}$

Mass of block B,$M_B=0.74\mathrm{kg}$

Coefficient of kinetic friction between the lighter block (Block A) and the plane,${\mu }_{K_A}=0.10$

Coefficient of kinetic friction between the heavier block (Block B) and the plane,${\mu }_{K_S}=0.20$

To find the acceleration of the blocks and the tension on the string, draw a free body diagram for each and then apply Newton's 2nd law of motion. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\sum _{}=ma$

where, F is the net force, mis mass and ais an acceleration.

Free Body Diagram of the Block A (lighter) and Block B (heavier):

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

$$\begin{gathered} N_A-W_A\cos \left(30\right)=0 \\ {N}_A-W_A\cos \left(30\right) \end{gathered}$$

Thus, the kinetic frictional force is,

$$\begin{gathered} f_{k_A}={\mu }_{k_A}{N}_A \\ ={\mu }_{k_A}{W}_A\cos \left(30\right) \end{gathered}$$

Similarly, to the horizontal direction, consider acceleration as a,

$$\begin{gathered} T+f_{k_A}-W_A\sin \left(30\right)=m_{A}a \\ T+{\mu }_{k_A}{W}_A\cos \left(30\right)-W_A\sin \left(30\right)=m_{A}a \end{gathered}$$

…(i)

Now, by using Newton’s 2nd law motion along the vertical and horizontal direction to the block B,

$$\begin{gathered} N_B-W_B\cos \left(30\right)=0 \\ {N}_B-W_B\cos \left(30\right)=0 \end{gathered}$$

Thus, the kinetic frictional force is,

$$\begin{gathered} f_{K_B}={\mu }_{K_B}{N}_B \\ ={\mu }_{K_B}{W}_B\cos \left(30\right) \end{gathered}$$

Along horizontal direction, consider acceleration is a,

localid="1657173983482" $$\begin{gathered} f_{K_B}T-W_B\sin \left(30\right)=m_{B}a \\ {\mu }_{K_B}{W}_{B}cos\left(30\right)-T-W_B\sin \left(30\right)=m_{B}a \end{gathered}$$

…(ii)

Adding equation (i) and (ii),

$$\begin{gathered} {\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\cos \left(30\right)-{\mathrm{W}}_{\mathrm{A}}\sin \left(30\right)+{\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\cos \left(30\right)-{\mathrm{W}}_{\mathrm{A}}\sin \left(30\right)=\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)\mathrm{a} \\ \left({\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{A}}}{\mathrm{W}}_{\mathrm{A}}+{\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\right)\cos \left(30\right)-\left({\mathrm{W}}_{\mathrm{A}}+{\mathrm{W}}_{\mathrm{B}}\right)\sin \left(30\right)=\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)\mathrm{a} \\ \mathrm{a}=\frac{\left({\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{A}}}{\mathrm{W}}_{\mathrm{A}}+{\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\right)}{\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)}\cos \left(30\right)-\mathrm{g}\sin \left(30\right) \\ =\left[\frac{\left(0.10\times 3.6\mathrm{N}+0.20\times 7.2\mathrm{N}\right)}{\left(0.37\mathrm{kg}+0.74\mathrm{kg}\right)}\cos \left(30\right)\right]-\left(9.8\mathrm{m}/{\mathrm{s}}^2\times \sin \left(30\right)\right) \\ =-3.5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Negative sign shows that blocks are moving to the downhill.

Thus,the magnitude of the acceleration is $3.5m/s^2$.

By using the expression of acceleration in equation (i),

$$\begin{gathered} \mathrm{T}+{\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{A}}}{\mathrm{W}}_{\mathrm{A}}\cos \left(30\right)-{\mathrm{W}}_{\mathrm{A}}\sin \left(30\right)={\mathrm{m}}_{\mathrm{A}}\left(\frac{\left({\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{A}}}{\mathrm{W}}_{\mathrm{A}}+{\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\right)}{\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)}\cos \left(30\right)-\mathrm{g}\sin \left(30\right)\right) \\ \mathrm{T}=\frac{{\mathrm{W}}_{\mathrm{A}}{\mathrm{W}}_{\mathrm{B}}}{\left({\mathrm{W}}_{\mathrm{A}}+{\mathrm{W}}_{\mathrm{B}}\right)}\left({\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{B}}}-{\mathrm{\mu }}_{{\mathrm{K}}_{\mathrm{A}}}\right)\cos \left(30\right) \\ =\frac{3.6\mathrm{N}\times 7.2\mathrm{N}}{\left(3.6\mathrm{N}+7.2\mathrm{N}\right)}\left(0.20-0.10\right)\cos \left(30\right) \\ =0.21\mathrm{N} \end{gathered}$$

Thus, the magnitude of tension on the string is 0.21 N.