Q92P - University Physics with Modern Physics

Question

Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg, which in turn is on a horizontal tabletop (Fig. P5.92). There is no friction between block A and the tabletop, but the coefficient of static friction between blocks A and B is 0.750. A light string attached to block A passes over a frictionless, massless pulley, and block C is suspended from the other end of the string. What is the largest mass that block C can have so that blocks A and B still slide together when the system is released from rest?

                                                           

                                         

Step-by-Step Solution

Answer

The largest mass the block C can have, so that there is no relative motion between block A and B is 39 kg.

Step 1: Identification of the given data

The given data can be listed below as:

• The mass of block B is $m_1=5.00\hspace{0.33em}kg$.
• The mass of block A is $m_2=8.00\hspace{0.33em}kg$.
• The static frictional coefficient between blocks A and B is ${\mu }_s=0.750$.

Step 2: Significance of the tension

Tension is described as a force that acts on a body by virtue of its elastic nature. Moreover, tension is also considered as a pair of action and reaction forces.

Step 3: Determination of the largest mass

The equation of the frictional force on the block B is expressed as:

 

$f={\mu }_s{m}_{1}g$ 

 

Here, f is the frictional force, ${\mu }_s$ is the static frictional coefficient between the block A and B, $m_1$ is the mass of the block B and is the acceleration due to gravity.

 

Substitute the values in the above equation.

 

$$\begin{gathered} f=\left(0.750\right)\left(5.00 kg\right)\left(9.8m/s^2\right) \\ =\left(3.75 kg\right)\left(9.8 m/s^2\right) \\ =36.75 kg·m/s^2\times \frac{1N}{1kg·m/s^2} \\ =36.75 N \end{gathered}$$

 

The equation of the acceleration of the block B is expressed as:

 

$a=\frac{f}{m_1}$ 

 

Here, a is the acceleration of the block B.

 

Substitute the values in the above equation.

 

$$\begin{gathered} a=\frac{36.75 N}{5.00 kg} \\ =7.35 N/kg \\ =7.35 N/kg\times \frac{ 1 kg·m/s^2}{1N} \\ =7.35 m/s^2 \end{gathered}$$

 

The equation of the mass of the block C is expressed as:

 

$$\begin{gathered} Mg-f=Ma+m_{2}a \\ {m}_{2}a+f=M\left(g-a\right) \\ \frac{m_{2}a+f}{g-a}=M \end{gathered}$$ 

 

Here, M is the mass of the block C.

 

Substitute the values in the above equation.

 

$$\begin{gathered} \mathrm{M}=\frac{\left(8.00 \mathrm{kg}\right)\left(7.35 \mathrm{m}/{\mathrm{s}}^2\right)+36.75 \mathrm{N}}{\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)-\left(7.35 \mathrm{m}/{\mathrm{s}}^2\right)} \\ =\frac{\left(58.8 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\right)+36.75 \mathrm{N}\times {\displaystyle \frac{1 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1 \mathrm{N}}}}{\left(2.45 \mathrm{m}/{\mathrm{s}}^2\right)} \\ =\frac{\left(95.55 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\right)}{\left(2.45 \mathrm{m}/{\mathrm{s}}^2\right)} \\ =39 \mathrm{kg} \end{gathered}$$

 

Thus, the largest mass the block C can have is 39 kg.