Q101P - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(a) The free body diagrams of blocks A and B are as follows.

And,

(b) The tension in the rope connecting blocks  A and B  is 8.75 N .

(c) The weight of the block  C is 30.75 N .

(d) The acceleration of the block  C is $1.54m/s^2$ .

Step 1: Identify the given data

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

Step 2: Concept/Significance of Kinetic friction force

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.

Step 3: Draw free-body diagrams for the forces acting on A and on B separately (a)

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.

Step 4: Find the tension in the rope connecting blocks A and B (b)

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 .

Step 5: Find the weight of block C (c)

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,

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

 

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

Step 6: Find the acceleration of C (d)

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:

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

 

Calculate the mass of lock B as:

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

 

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,

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

 

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