Chapter 12.1, Problem 12.18P

Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are ${\mu }_s=0.20$ and ${\mu }_k=0.15$ . If $P=0$ , determine (a) the acceleration of block B, (b) the tension in the cord.

  

To determine

(a)

Acceleration of block B

Answer to Problem 12.18P

Acceleration of block B:

$a_B=0.986\text{m}/s^2$

Explanation of Solution

Given:

$m_A=40\text{kg,}{m}_B=8\text{kg}$

${\mu }_s=0.20,{\mu }_k=0.15$

$P=0$

From the constraint of the cord:

$2x_A+x_{B/A}=\text{constant}$

Then

$2v_A+v_{B/A}=0$

And

$2a_A+a_{B/A}=0$

Now

$a_B=a_A+a_{B/A}$

Then

$a_B=a_A+\left(-2a_A\right)$

Or:

$a_B=-a_A$   ... (1)

First, we determine if the blocks will move for the given value of $\theta$. Thus, we seek the value of $\theta$ for which the blocks are in impeding motion, with the impeding motion of A down the incline.

For B:

$\nearrow \sum F_y=0$ :

$N_{AB}-W_B\cos \theta =0$

Or:

$N_{AB}=W_B\cos \theta$

Now

$F_{AB}={\mu }_s{N}_{AB}$

$F_{AB}=0.2m_{B}g\cos \theta$

${\searrow }^{+}\sum F_x=0$

$-T+F_{AB}+W_B\sin \theta =0$

Or:

$T=m_{B}g\left(0.2\cos \theta +W_B\sin \theta \right)$

For A:

$\nearrow \sum F_y=0$ :

$N_A-N_{AB}-W_A\cos \theta =0$

Or:

$N_A=\left(m_A+m_B\right)g\cos \theta$

Now

$F_A={\mu }_s{N}_A$$F_A=0.2\left(m_A+m_B\right)g\cos \theta$

${\searrow }^{+}\sum F_x=0$ :

$-T-F_A-F_{AB}+W_A\sin \theta =0$

$T=m_{A}g\sin \theta -0.2\left(m_A+m_B\right)g\cos \theta -0.2m_{B}g\cos \theta$

$T=g\left[m_A\sin \theta -0.2\left(m_A+m_B\right)g\cos \theta \right]$

Equating the two expressions for T,

$m_{B}g\left(0.2\cos \theta +\sin \theta \right)=g\left[m_A\sin \theta -0.2\left(m_A+m_B\right)g\cos \theta \right]$

Or:

$8\left(0.2+\tan \theta \right)=g\left[40\tan \theta -0.2\left(40+2\times 8\right)\right]$

Or:

$\tan \theta =0.4$

Or:

$\theta ={21.8}^{\circ }$, for impeding motion

Since $\theta <{25}^{\circ }$, the blocks will move.

Now, consider the motion of the blocks.

For B:

$\nearrow \sum F_y=0$ :

$N_{AB}-W_B\cos {25}^{\circ }=0$

Or:

$N_{AB}=m_{B}g\cos {25}^{\circ }$

Sliding:

$F_{AB}={\mu }_k{N}_{AB}=0.15m_B\cos {25}^{\circ }$

${\searrow }^{+}\sum F_x=m_B{a}_B$

$-T+F_{AB}+W_B\sin {25}^{\circ }=m_B{a}_B$

Or:

$T=m_B\left[g\left(0.15\cos {25}^{\circ }+\sin {25}^{\circ }\right)-a_B\right]$

$T=8\left[9.81\left(0.15\cos {25}^{\circ }+\sin {25}^{\circ }\right)-a_B\right]$

$T=8\left(5.47952-a_B\right)$

For A:

$\nearrow \sum F_y=0$ :

$N_A-N_{AB}-W_A\cos {25}^{\circ }=0$

Or:

$N_A=\left(m_A+m_B\right)g\cos {25}^{\circ }$

Sliding:

$F_A={\mu }_k{N}_A=0.15\left(m_A+m_B\right)g\cos {25}^{\circ }$

${\searrow }^{+}\sum F_x=m_A{a}_A$ :

$-T-F_A-F_{AB}+W_A\sin {25}^{\circ }=m_A{a}_A$

Substituting and using Eq. (1):

$T=m_{A}g\sin {25}^{\circ }-0.15\left(m_A+m_B\right)g\cos {25}^{\circ }-0.15m_{B}g\cos {25}^{\circ }-m_A\left(-a_B\right)$

$T=g\left[m_A\sin {25}^{\circ }-0.15\left(m_A+m_B\right)\cos {25}^{\circ }\right]+m_A{a}_B$

$T=9.81\left[40\sin {25}^{\circ }-0.15\left(40+2\times 8\right)\cos {25}^{\circ }\right]+40a_B$

$T=91.15202+40a_B$

Equating the two expressions for T,

$8\left(5.47952-a_B\right)=91.15202+40a_B$

Or:

$a_B=-0.98575\text{m}/s^2$

$a_B=0.986\text{m}/s^2$

To determine

(b)

Tension in the cable

Answer to Problem 12.18P

Tension in the cable:

$T=51.7\text{N}$

Explanation of Solution

Given:

$m_A=40\text{kg,}{m}_B=8\text{kg}$

${\mu }_s=0.20,{\mu }_k=0.15$

$P=0$

$T=8\left(5.47952-a_B\right)$

$T=8\left[5.47952-\left(-0.98575\right)\right]$

Or:

$T=51.7\text{N}$