Question

Two wooden crates rest on top of one another. The smaller top crate has a mass of M1=22kg and the larger bottom crate has a mass of M2=92kg. There is No friction between the crate and the floor, but the coefficient of static friction between the two crates is Ms=0.87 and the coefficient of kinetic friction between the two crates is MK=0.68. A massless rope is attached to the lower crate to pull it horizontally to the right. (Which should be considered the positive direction for this problem). 1) the rope is pulled with a tension of T = 500N. What is the acceleration of the smaller crate?.2) In the previous situation, what is the frictional force the lower crate exerts on the upper crate?3) What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4) The tension is increased in the rope to 1440N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?5) As the upper crate slides, what is the acceleration of the lower crate?

Answer 1

`\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}`

Step-by-step 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:

`F=m_1a_1`

The sum of forces acting in the horizontal direction on the larger crate is:

`T-F=m_2a_2`

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:

`T-m_1a_1=m_2a_2`

The acceleration is equal for both crates since the upper crate is static during the motion, which implies that:

`\begin{gathered} T=m_1a+m_2a \\ T=a(m_1+m_2) \\ \Rightarrow a=(T)/(m_1+m_2) \end{gathered}`

Substitute the given values:

`\begin{gathered} a=(500)/(22+92) \\ a=4.39m/s^2 \end{gathered}`

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:

`\begin{gathered} \Rightarrow F=22\cdot4.39 \\ F=96.58N \end{gathered}`

3) The maximum tension that the rope can be pulled at will occur when there is maximum friction.

That is:

`F_{s(\text{max)}}=\mu_s\cdot m_2\cdot g`

Applying the same principle from the first equation:

`\begin{gathered} F=ma \\ \Rightarrow F_2=m_2a \end{gathered}`

This implies that:

`\begin{gathered} m_2a=\mu_s\cdot m_2\cdot g \\ \Rightarrow a=\mu_s\cdot g \end{gathered}`

To find the tension, apply the already given formula:

`\begin{gathered} T=m_1a+m_2a=(m_1+m_2)a \\ T=(m_1+m_2)\cdot\mu_s\cdot g \end{gathered}`

Therefore, the max tension is:

`\begin{gathered} T=(22+92)\cdot0.87\cdot9.8 \\ T=114\cdot0.87\cdot9.8 \\ T=971.96N \end{gathered}`

4) To find the acceleration of the smaller crate, we apply the formula for kinetic friction:

`F_k=\mu_k\cdot m_1\cdot g`

Recall that:

`F=m_1\cdot a_1`

Therefore:

`\begin{gathered} m_1\cdot a_1=\mu_k\cdot m_1\cdot g \\ \Rightarrow a_1=\mu_k\cdot g \end{gathered}`

Substitute the given values:

`\begin{gathered} a_1=0.68\cdot9.8 \\ a_1=6.66m/s^2 \end{gathered}`

That is the acceleration of the smaller/upper crate.

5) Recall from the second equation that:

`T-F=m_2a_2`

Substitute the formula for kinetic friction:

`\begin{gathered} T-\mu_k\cdot m_1\cdot g=m_2\cdot a_2 \\ \Rightarrow a_2=(T-\mu_k\cdot m_1\cdot g)/(m_2) \end{gathered}`

Substitute the given values to solve for a2:

`\begin{gathered} a_2=(1440-(0.68\cdot22\cdot9.8))/(92) \\ a_2=(1440-146.608)/(92) \\ a_2=14.06m/s^2 \end{gathered}`

That is the acceleration of the larger/lower crate.