Chapter 4, Problem 90P

To determine

Tension in the rope between the crates and the magnitude of the acceleration of the system.

Answer to Problem 90P

Tension in the rope between the crates is $86\text{N}$ and the magnitude of the acceleration of the system is $4.2{\text{m/s}}^2$.

Explanation of Solution

Figure 1 shows the free body diagram of the system.

Write an expression to calculate the net horizontal force on crate 1.

$\begin{array}{c}\Sigma F_{1x}=T-f_1\\ =T-\mu N_1\\ =m_{1}a\end{array}$ (I)

Here, the net horizontal force on crate 1 is $\Sigma F_{1x}$, the tension on the rope is $T$, the frictional force on crate 1 is $f_1$, the coefficient of kinetic friction is $\mu$, the normal force on crate 1 is $N_1$, the mass of crate 1 is $m_1$ and the acceleration is $a$.

Rearrange the equation (I) to find $T$.

$T=m_{1}a+\mu N_1$ (II)

Write an expression to calculate the net vertical force on crate 1.

$\begin{array}{c}\Sigma F_{1y}=N_1-m_{1}g\\ =0\end{array}$ (III)

Here, the net vertical force on crate 1 is $\Sigma F_{1y}$.

Rearrange the equation (III) to find $N_1$.

$N_1=m_{1}g$ (IV)

Substitute equation (IV) in (II)

$\begin{array}{c}T=m_{1}a+\mu \left(N_1\right)\\ =m_{1}a+\mu m_{1}g\\ =m_1\left(a+\mu g\right)\end{array}$ (V)

Write an expression to calculate the net horizontal force on crate 2.

$\begin{array}{c}\Sigma F_{2x}=F\cos \theta -T-f_2\\ =F\cos \theta -\left(m_{1}a+\mu m_{1}g\right)-\mu N_2\\ =m_{2}a\end{array}$ (VI)

Here, the net horizontal force on crate 2 is $\Sigma F_{2x}$, the frictional force on crate 2 is $f_2$, the forward force is $F$, the angle at which the forward force acting is $\theta$, the mass of crate 2 is $m_2$ and the normal force on crate 2 is $N_2$.

Write an expression to calculate the net vertical force on crate 2.

$\begin{array}{c}\Sigma F_{2y}=N_2+F\sin \theta -m_{2}g\\ =0\end{array}$ (VII)

Here, the net vertical force on crate 2 is $\Sigma F_{2y}$.

Rearrange the equation (VII) to find $N_2$.

$N_2=m_{2}g+F\sin \theta$ (VIII)

Substitute equation (VIII) in (VI)

$F\cos \theta -\left(m_{1}a+\mu m_{1}g\right)-\mu \left(m_{2}g+F\sin \theta \right)=m_{2}a$ (IX)

Rearrange the equation (IX) to find $a$.

$a=\frac{F\left(\cos \theta +\mu \sin \theta \right)}{\left(m_1+m_2\right)}-\mu g$ (X)

Conclusion:

Substitute $195\text{N}$ for $F$, $20.0°$ for $\theta$, $0.550$ for $\mu$, $9.00\text{kg}$ for $m_1$, $14.0\text{kg}$ for $m_2$, and $9.80{\text{m/s}}^2$ for $g$ in equation (X) to find $a$.

$\begin{array}{c}a=\frac{\left(195\text{N}\right)\left(\cos \left(20.0°\right)+\left(0.550\right)\sin \left(20.0°\right)\right)}{\left(9.00\text{kg}+14.0\text{kg}\right)}-\left(0.550\right)\left(9.80{\text{m/s}}^2\right)\\ =\frac{\left(195\text{N}\right)\left(\cos \left(20.0°\right)+\left(0.550\right)\sin \left(20.0°\right)\right)}{\left(25.0\text{kg}\right)}-\left(0.550\right)\left(9.80{\text{m/s}}^2\right)\\ =4.2{\text{m/s}}^2\end{array}$

Substitute $9.00\text{kg}$ for $m_1$, $0.550$ for $\mu$, $9.80{\text{m/s}}^2$ for $g$, and $4.2{\text{m/s}}^2$ for $a$ in equation (V) to find $T$.

$\begin{array}{c}T=\left(9.00\text{kg}\right)\left(\left(4.2{\text{m/s}}^2\right)+\left(0.550\right)\left(9.80{\text{m/s}}^2\right)\right)\\ =\left(9.00\text{kg}\right)\left(9.59\text{kg}\right)\\ =86\text{N}\end{array}$

Thus, the tension in the rope between the crates is $86\text{N}$ and the magnitude of the acceleration of the system is $4.2{\text{m/s}}^2$.