ANSWER
[tex]\begin{gathered} 1)4.39m/s^2 \\ 2)96.58N \\ 3)971.96N \\ 4)6.66m/s^2 \\ 5)14.06m/s^2 \end{gathered}[/tex]
EXPLANATION
Parameters given:
Mass of smaller crate, m1 = 22 kg
Mass of larger crate, m2 = 92 kg
Coefficient of static friction between two crates, μs = 0.87
Coefficient of kinetic friction between two crates, μk = 0.68
1) The sum of forces acting in the horizontal direction on the smaller crate is:
[tex]F=m_1a_1[/tex]
The sum of forces acting in the horizontal direction on the larger crate is:
[tex]T-F=m_2a_2[/tex]
where F = frictional force, T = tension
To find the acceleration of the smaller crate, we have to substitute F from the first equation into the second equation:
[tex]T-m_1a_1=m_2a_2[/tex]
The acceleration is equal for both crates since the upper crate is static during the motion, which implies that:
[tex]\begin{gathered} T=m_1a+m_2a \\ T=a(m_1+m_2) \\ \Rightarrow a=\frac{T}{m_1+m_2} \end{gathered}[/tex]
Substitute the given values:
[tex]\begin{gathered} a=\frac{500}{22+92} \\ a=4.39m/s^2 \end{gathered}[/tex]
That is the acceleration of the smaller crate.
2) To find the frictional force on the smaller crate, substitute the value of acceleration, a, into the first equation:
[tex]\begin{gathered} \Rightarrow F=22\cdot4.39 \\ F=96.58N \end{gathered}[/tex]
3) The maximum tension that the rope can be pulled at will occur when there is maximum friction.
That is:
[tex]F_{s(\text{max)}}=\mu_s\cdot m_2\cdot g[/tex]
Applying the same principle from the first equation:
[tex]\begin{gathered} F=ma \\ \Rightarrow F_2=m_2a \end{gathered}[/tex]
This implies that:
[tex]\begin{gathered} m_2a=\mu_s\cdot m_2\cdot g \\ \Rightarrow a=\mu_s\cdot g \end{gathered}[/tex]
To find the tension, apply the already given formula:
[tex]\begin{gathered} T=m_1a+m_2a=(m_1+m_2)a \\ T=(m_1+m_2)\cdot\mu_s\cdot g \end{gathered}[/tex]
Therefore, the max tension is:
[tex]\begin{gathered} T=(22+92)\cdot0.87\cdot9.8 \\ T=114\cdot0.87\cdot9.8 \\ T=971.96N \end{gathered}[/tex]
4) To find the acceleration of the smaller crate, we apply the formula for kinetic friction:
[tex]F_k=\mu_k\cdot m_1\cdot g[/tex]
Recall that:
[tex]F=m_1\cdot a_1[/tex]
Therefore:
[tex]\begin{gathered} m_1\cdot a_1=\mu_k\cdot m_1\cdot g \\ \Rightarrow a_1=\mu_k\cdot g \end{gathered}[/tex]
Substitute the given values:
[tex]\begin{gathered} a_1=0.68\cdot9.8 \\ a_1=6.66m/s^2 \end{gathered}[/tex]
That is the acceleration of the smaller/upper crate.
5) Recall from the second equation that:
[tex]T-F=m_2a_2[/tex]
Substitute the formula for kinetic friction:
[tex]\begin{gathered} T-\mu_k\cdot m_1\cdot g=m_2\cdot a_2 \\ \Rightarrow a_2=\frac{T-\mu_k\cdot m_1\cdot g}{m_2} \end{gathered}[/tex]
Substitute the given values to solve for a2:
[tex]\begin{gathered} a_2=\frac{1440-(0.68\cdot22\cdot9.8)}{92} \\ a_2=\frac{1440-146.608}{92} \\ a_2=14.06m/s^2 \end{gathered}[/tex]
That is the acceleration of the larger/lower crate.
