Chapter 8, Problem 8.134RP

8.134 and 8.135 The coefficients of friction are μ_S = 0.40 and μ_k = 0.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as shown, (b) is removed.

To determine

Find the smallest value of P required to start moving the 30 kg block if the cable AB is attached.

Answer to Problem 8.134RP

The smallest force P required to move the 30-kg block is $\underset{\_}{353\text{N}(\leftarrow )}$.

Explanation of Solution

Given information:

The coefficient of static friction is ${\mu }_s=0.40$.

The coefficient of kinetic friction is ${\mu }_k=0.30$.

Calculation:

Consider the acceleration due to gravity as $g=9.81{\text{m/s}}^2$.

Find the weight of the 20-kg mass block as follows;

$\begin{array}{c}{W}_{\text{20-kg}}=m_{20}g\\ =20\times 9.81\\ =196.2\text{N}\end{array}$

Find the weight of the 30-kg mass block as follows;

$\begin{array}{c}{W}_{\text{30-kg}}=m_{30}g\\ =30\times 9.81\\ =294.3\text{N}\end{array}$

Show the free-body diagram of the 20-kg mass block as in Figure 1.

Find the normal force $\left(N_{20}\right)$ by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ N_{20}-W_{20}=0\\ N_{20}=196.2\text{N}\end{array}$

Find the friction force $\left(F_{20}\right)$ using the relation.

$F_{20}={\mu }_s{N}_{20}$

Substitute 0.40 for ${\mu }_s$ and 196.2 N for $N_{20}$.

$\begin{array}{c}{F}_{20}=0.40\times 196.2\\ =78.48\text{N}\end{array}$

Find the tension in the cable AB (T) by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ T-F_{20}=0\\ T=78.48\text{N}\end{array}$

Show the free-body diagram of the 30-kg mass block as in Figure 2.

Find the normal force $\left(N_{30}\right)$ by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -N_{20}-W_{30}+N_{30}=0\\ -196.2-294.3+N_{30}=0\\ N_{30}=490.5\text{N}\end{array}$

Find the friction force $\left(F_{30}\right)$ using the relation.

$F_{30}={\mu }_s{N}_{30}$

Substitute 0.40 for ${\mu }_s$ and 490.5 N for $N_{30}$.

$\begin{array}{c}{F}_{30}=0.40\times 490.5\\ =196.2\text{N}\end{array}$

Find the force P by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -P+T+F_{20}+F_{30}=0\\ -P+78.48+78.48+196.2=0\\ P=353\text{N}(\leftarrow )\end{array}$

Therefore, the smallest force P required to move the 30-kg block is $\underset{\_}{353\text{N}(\leftarrow )}$.

To determine

Find the smallest value of P required to start moving the 30 kg block if the cable is removed.

Answer to Problem 8.134RP

The smallest force P required to move the 30-kg block is $\underset{\_}{196.2\text{N}(\leftarrow )}$.

Explanation of Solution

Given information:

The coefficient of static friction is ${\mu }_s=0.40$.

The coefficient of kinetic friction is ${\mu }_k=0.30$.

Calculation:

Consider the acceleration due to gravity as $g=9.81{\text{m/s}}^2$.

Find the weight of the 20-kg mass block as follows;

$\begin{array}{c}{W}_{\text{20-kg}}=m_{20}g\\ =20\times 9.81\\ =196.2\text{N}\end{array}$

Find the weight of the 30-kg mass block as follows;

$\begin{array}{c}{W}_{\text{30-kg}}=m_{30}g\\ =30\times 9.81\\ =294.3\text{N}\end{array}$

Show the free-body diagram of the block assembly as in Figure 3.

Find the normal force (N) by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ N-W_{20}-W_{30}=0\\ N-196.2-294.3=0\\ N=490.5\text{N}\end{array}$

Find the friction force (F) using the relation.

$F={\mu }_{s}N$

Substitute 0.40 for ${\mu }_s$ and 490.5 N for N.

$\begin{array}{c}F=0.40\times 490.5\\ =196.2\text{N}\end{array}$

Find the force P by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -P+F=0\\ P=196.2\text{N}(\leftarrow )\end{array}$

Therefore, the smallest force P required to move the 30-kg block is $\underset{\_}{196.2\text{N}(\leftarrow )}$.