Blocks A, B, and C are placed as in Fig. P5.101 and connected by ropes of negligible mass. Both A and B weigh 25.0 N each, and the coefficient of kinetic friction between each block and the surface is 0.35. Block C descends with constant velocity.

(a) Draw separate free-body diagrams showing the forces acting on A and on B.

(b) Find the tension in the rope connecting blocks A and B.

(c) What is the weight of block C?

(d) If the rope connecting A and B were cut, what would be the acceleration of C?

• The weight of blocks A and B, $w=25N$.
• The coefficient of kinetic friction, ${\mu }_k=0.35$.

The kinetic friction force occurs when the object is in motion. The kinetic friction force is given by,

${\mathit{f}}_{\mathbf{k}}\mathbf{=}{\mathit{\mu }}_{\mathbf{k}}\mathit{N}$

Here, ${\mathit{\mu }}_{\mathbf{k}}$ is the coefficient of kinetic friction, and N is normal force.

Draw the free-body diagram for block A.

Here, $T_1$is the tension in the rope between blocks A and B , $f_1$is the frictional force between the block A and the surface, and $N_1$ is the normal reaction between the block A and the surface.

Draw the free-body diagram for block B.

Here, $T_1$ is the tension in the rope between blocks A and B , $T_2$ is the tension in the rope between blocks B and C , n is the normal reaction on the block B , and f is the frictional force between the block B and the surface.

The force along the y-axis for the block A is given by,

$N_1-w=0$

The force along the x-axis for the block A is given by,

$T_1-f_1=0$

The frictional force between the block A and the surface is given by,

$$\begin{gathered} f_1={\mu }_k{N}_1 \\ ={\mu }_{k}w \end{gathered}$$

Substitute the frictional force expression in the equation $T_1-f_1=0$, and we get,

$$\begin{gathered} T_1-{\mu }_{k}w=0 \\ {T}_1={\mu }_{k}w.......\left(1\right) \end{gathered}$$

Substitute 0.35 for ${\mu }_k$ and 25 N for w in equation (1), and we get,

$$\begin{gathered} T_1=\left(0.35\right)\left(25N\right) \\ =8.75N \end{gathered}$$

Therefore, the tension in the rope connecting blocks A and B is 8.75 N .

Draw the free-body diagram for block C.

Here, the weight of the block C is $w_c$.

The force equation for the block C is given by,

$w_c-T_2=0.......\left(2\right)$

The force along the x-axis for the block B is given by,

$T_2-T_1-f-w\sin \theta =0........\left(3\right)$

The force along the y-axis for the block B is given by,

$N=w\cos \theta ......\left(4\right)$

The frictional force between the block B and the surface is given by,

$f={\mu }_{k}N$

From equation (4),

$f={\mu }_k\left(w\cos \theta \right).....\left(5\right)$

Noting that block C descends down at constant velocity, so its acceleration is zero.

$$\begin{gathered} w_c-T_2=0 \\ {w}_c=T_2 \end{gathered}$$

Substitute equation (5) in (3), and we get,

$$\begin{gathered} T_2-T_1-{\mu }_k\left(w\cos \theta \right)-w\sin \theta =0 \\ {w}_c=T_1+{\mu }_k\left(w\cos \theta \right)+w\sin \theta .....\left(6\right) \end{gathered}$$

Substitute 8.75 N for $T_1$, $36.9°$for $\theta$, 0.35 for ${\mu }_k$, and 25 N for w in equation (6), and we get,

$$\begin{gathered} w_c=8.75N+\left(0.35\right)\left(25N\right)\cos \left(36.9°\right)+\left(25N\right)\sin \left(36.9°\right) \\ =30.75N \end{gathered}$$

Therefore, the weight of the block C is 30.75N .

Consider being the acceleration of both the masses B and C after the rope is cut, and T is the new tension between blocks B and C .

Draw the free-body diagram of the block B after the rope is cut between A and B .

The force along the x-axis for the block B is given by,

$T-f-w\sin \theta =m_{B}a..........\left(7\right)$

The force along the y-axis for the block B is given by,

$N-w\sin \theta =0.......\left(8\right)$

The frictional force between the block B and the surface is given by,

$f={\mu }_{k}N........\left(9\right)$

Substitute equations (8) and (9) in the equation (7), and we get,

$T-{\mu }_{k}w\sin \theta -w\sin \theta =m_{B}a........\left(10\right)$

Draw the free-body diagram of the block C after the rope is cut between A and B .

The new force for the block C is given by,

$w_c-T=m_{c}a........\left(11\right)$

From equations (10) and (11),

$a=\frac{w_c-{\mu }_k\left(w\cos \theta \right)-w\sin \theta }{m_c+m_B}........\left(12\right)$

Calculate the mass of lock C as:

$$\begin{gathered} m_c=\frac{w_c}{g} \\ =\frac{30.75N}{9.8m/s^2} \\ =3.14kg \end{gathered}$$

Calculate the mass of lock B as:

$$\begin{gathered} m_B=\frac{w_B}{g} \\ =\frac{25N}{9.8m/s^2} \\ =2.55kg \end{gathered}$$

Substitute 30.75N for $w_c$, 0.35 for ${\mu }_k$, 25 N for w , 3.14 kg for $m_c$, 2.55kg for $m_B$ , and $36.9°$ for $\theta$ in equation (12), and we get,

$$\begin{gathered} a=\frac{30.75N-\left(0.35\right)\left(25N\right)\cos \left(36.9°\right)-\left(25N\right)\sin \left(36.9°\right)}{3.14kg+2.55kg} \\ =1.54m/s^2 \end{gathered}$$

Therefore, the acceleration of the block C is $1.54m/s^2$.