Q97P - University Physics with Modern Physics

Question

Block A, with weight 3w, slides down an inclined plane S of slope angle 36.9° at a constant speed while plank B, with weight w, rests on top of A. The plank is attached by a cord to the wall (Fig. P5.97). 

(a) Draw a diagram of all the forces acting on block A. 

(b) If the coefficient of kinetic friction is the same between A and B and between S and A, determine its value.

Step-by-Step Solution

Answer

(a) The diagram of all forces on the block  A is as follows.

(b) The value of the coefficient of kinetic friction is 0.451 .

Step 1: Describe Newton’s second law and Kinetic friction force

According to Newton’s second law, the linear force is given by,

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$  

Here, F  is linear force,  m is the mass of the object, and a  is the acceleration of the object.

 

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   is the normal force.

 

Given data:

• slope $\theta =36.9°$ .

Step 2: Draw a diagram of all the forces acting on block A (a)

The free-body diagram of the block  A is shown below.

From the above figure, it is obtained that the normal force  $F_n$ acting along the upward direction, the force due to gravity 3w  acting along the downward direction. The components of force due to gravity are $3w \sin \theta$  and $3w \cos \theta$ . The force $f_A$  is the force on top of A .

Step 3: Find the coefficient of kinetic friction (b)

Let the coefficient of kinetic friction between A  and B , and between  S and  A be ${\mu }_k$ .

 

The frictional force between  S and  A is given by,

  $$\begin{gathered} f_A={\mu }_k\left(F_n\right) \\ ={\mu }_k\left(4w\cos \theta \right) \end{gathered}$$

 

The normal force between the block and the plank is given by,

  $F_{n_1}=w \cos \theta$

 

The normal force between the block and the inclined plane is given by,

$F_{n_2}=4w \cos \theta$  

 

The net normal force between the block and the plank is given by,

 $$\begin{gathered} F_n=F_{n_1}+F_{n_2} \\ =w \cos \theta +4w \cos \theta \\ =5w \cos \theta \end{gathered}$$ 

 

The frictional force between  A and  B is given by,

$f_B={\mu }_k\left(w\cos \theta \right)$  

 

Since A  is sliding down at constant speed, so calculate the net force on the block  A by using Newton’s second law.

 

$\sum _{ }{F}_x=ma$  

 

Here  $m_A$ is the mass and  a is the acceleration.

 

Substitute $3w \sin \theta -5w {\mu }_k\cos \theta$  for $\sum _{ }{F}_x$ , and  $36.9°$ for  $\theta$ in the above equation.

  $$\begin{gathered} 3w \sin \theta -5w {\mu }_k\cos \theta =0 \\ {\mu }_k=\frac{3}{5}\tan \theta \\ =\frac{3}{5}\tan \left(36.9°\right) \\ {\mu }_k=0.451 \end{gathered}$$

 

Therefore, the value of the coefficient of kinetic friction is  0.451.