Q5-87P - University Physics with Modern Physics

Question

Block A in Fig. P5.87 weighs $1.90\text{\hspace{0.17em}N}$, and block B weighs .$4.20\text{\hspace{0.17em}N}$ The coefficient of kinetic friction between all surfaces is $0.30$. Find the magnitude of the horizontal force $\overrightarrow{F}$ necessary to drag block B to the left at constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley.

Step-by-Step Solution

Answer

The magnitude of the horizontal $\overrightarrow{F}$ force is $3\text{\hspace{0.17em}N}$.

Step 1: Identification of given data

The given data can be listed below as:

• The weight of the block A is .$W_1=1.90\text{\hspace{0.17em}N}$
• The weight of the block B is .$W_2=4.20\text{\hspace{0.17em}N}$
• The coefficient of the kinetic friction is .${\mu }_k=0.30$

Step 2: Significance of the friction

The friction is described as the force which opposes the motion of a particular object. The friction is also described as the pair of the action and the reaction forces.

Step 3: Determination of the horizontal force

The free body diagram of the system has been drawn below:

For the block B, the equation of the horizontal force exerted by the block B is expressed as:

 

                                  $F=f+T+f_c$                                                                         …(1)

 

Here,$F$ is described as the horizontal force,$f$ is the upper frictional force,$T$ is the tension of the block B and $f_c$ is the lower frictional force.

 

The equation of the upper frictional force is expressed as:

$f={\mu }_k{W}_1$

Here,${\mu }_k$ is the coefficient of the kinetic friction and $W_1$ is the weight of the block A.

 

According to the free body diagram of the block A, the tension exerted by the block is the upper frictional force.

 

The equation of the lower frictional force is expressed as:

 $f_c={\mu }_k(W_1+W_2)$

Here,$f_c$ is the lower frictional force, ${\mu }_k$ is the coefficient of the kinetic friction and $W_2$ is the weight of the block B.

 

Substitute ${\mu }_k(W_1+W_2)$for $f$ and $T$ and ${\mu }_k{W}_1$ for $f_c$ in the above equation.

 $\begin{array}{c}F={\mu }_k(W_1+W_2)+{\mu }_k(W_1+W_2)+{\mu }_k{W}_1\\ =2{\mu }_k(W_1+W_2)+{\mu }_k{W}_1\\ ={\mu }_k(3W_1+W_2)\end{array}$

Substitute the values in the above equation.

 $\begin{array}{c}F=\left(0.30\right)(3\left(1.90\text{\hspace{0.17em}N}\right)+\left(4.20\text{\hspace{0.17em}N}\right))\\ =\left(0.30\right)(\left(5.7\text{\hspace{0.17em}N}\right)+\left(4.20\text{\hspace{0.17em}N}\right))\\ =\left(0.30\right)\left(9.9\text{\hspace{0.17em}N}\right)\\ =2.97\text{\hspace{0.17em}N}\\ \approx \text{3\hspace{0.17em}N}\end{array}$

 

Thus, the magnitude of the horizontal force $\overrightarrow{F}$ is $3\text{\hspace{0.17em}N}$.