Chapter 4, Problem 65P

To determine

(a) The magnitude of the acceleration of $m_a\text{ and }{m}_b$ .(b)The smallest value of ${\mu }_k$ that will keep the system from accelerating.

Answer to Problem 65P

Solution:

  (a) Acceleration $a=1.55{\text{m/s}}^2$

  (b) ${\mu }_k=0.53$

Explanation of Solution

Given:

  $m_a=m_b=2.7\text{kg}$

  θ =34^o.

  ${\mu }_k=0.15$

Formula used:

From Newton’s second law of motion:

  ${\displaystyle \sum F=ma}$

Where, m is the mass and a is the acceleration.

Calculation:

  

For Block A:

  $\begin{array}{l}{m}_{a}g\sin \theta -{\mu }_{a}mg\cos \theta -T=m_{a}a\\ T-(2.7)(9.8)\sin 34-0.15(2.7)(9.8)\cos 34=2.7a\\ T-18.08=2.7a----------(1)\end{array}$

For Block B:

  $\begin{array}{l}{m}_{b}g-T=m_{b}a\\ 2.7(9.8)-T=2.7a\\ 26.46-T=2.7a----------(2)\end{array}$

Add equations (1) and (2):

  $\begin{array}{l}T-18.08+26.46-T=2.7a+2.7a\\ 8.38=5.4a\\ a=1.55{\text{m/s}}^2\end{array}$

(b) Acceleration = a= 0

For Block A:

  $\begin{array}{l}{m}_{a}g\sin \theta -{\mu }_{k}mg\cos \theta -T=m_{a}a\\ T-(2.7)(9.8)\sin 34-{\mu }_k(2.7)(9.8)\cos 34=2.7(0)\\ T-14.8-21.94{\mu }_k=0----------(1)\end{array}$

For Block B:

  $\begin{array}{l}{m}_{b}g-T=m_{b}a\\ 2.7(9.8)-T=2.7(0)\\ 26.46-T=0----------(2)\end{array}$

Add equations (1) and (2):

  $\begin{array}{l}T-14.8-21.94{\mu }_k+26.46-T=0\\ 11.66-21.94{\mu }_k=0\\ {\mu }_k=\frac{11.66}{21.94}\\ {\mu }_k=0.53\end{array}$

Conclusion:

Thus, the acceleration of the blocks is 1.55m/s² and the smallest value of ${\mu }_k$ required for no acceleration is 0.53.