Box on a Surface (Alternate Method)

Example: Newton's Second Law: Box on a Surface (Alternate Method)

Question

Two crates, $\text{10}$ $\text{kg}$ and $\text{15}$ $\text{kg}$ respectively, are connected with a thick rope according to the diagram. A force, to the right, of $\text{500}$ $\text{N}$ is applied. The boxes move with an acceleration of $\text{2}$ $\text{m·s$^{-2}$}$ to the right. One third of the total frictional force is acting on the $\text{10}$ $\text{kg}$ block and two thirds on the $\text{15}$ $\text{kg}$ block. Calculate:

the magnitude and direction of the total frictional force present.
the magnitude of the tension in the rope at T.

Tip:

Important: when you have tension in a rope in a problem like this you need to know that both ends of the rope apply a force with the same magnitude but opposite direction. We call this force the tension and you should examine the force diagrams in this problem carefully.

Step 1: Draw a force diagram

Always draw a force diagram although the question might not ask for it. The acceleration of the whole system is given, therefore a force diagram of the whole system will be drawn. Because the two crates are seen as a unit, the force diagram will look like this:

Step 2: Calculate the frictional force

To find the frictional force we will apply Newton's second law. We are given the mass ($\text{10}$ + $\text{15}$ $\text{kg}$) and the acceleration ($\text{2}$ $\text{m·s$^{-2}$}$). Choose the direction of motion to be the positive direction (to the right is positive).

\begin{align*} {F}_{R} & = ma\\ {F}_{\text{applied}}+{F}_{f} & =ma\\ \text{500}+{F}_{f} & =\left(10+15\right)\left(2\right)\\ {F}_{f} & =50-\text{500}\\ {F}_{f} & =-\text{450}N \end{align*}

The frictional force is $\text{450}$ $\text{N}$ opposite to the direction of motion (to the left).

Step 3: Find the tension in the rope

To find the tension in the rope we need to look at one of the two crates on their own. Let's choose the $\text{10}$ $\text{kg}$ crate. Firstly, we need to draw a force diagram:

Force diagram of $\text{10}$ $\text{kg}$ crate.

The frictional force on the $\text{10}$ $\text{kg}$ block is one third of the total, therefore:

${F}_{f}=\frac{\text{1}}{\text{3}}\times \text{450}$

${F}_{f}=\text{150} \text{N}$

If we apply Newton's second law:

\begin{align*} {F}_{R}& = ma \\ T+{F}_{f}& = \left(10\right)\left(2\right) \\ T+\left(-\text{150}\right)& = 20 \\ T& = \text{170}\text{ N} \end{align*}

Note: If we had used the same principle and applied it to $\text{15}$ $\text{kg}$ crate, our calculations would have been the following:

\begin{align*} {F}_{R}& = ma \\ {F}_{\text{applied}}+T+{F}_{f}& = \left(15\right)\left(2\right) \\ \text{500}+T+\left(-\text{300}\right)& = 30 \\ T& = -\text{170}\text{ N} \end{align*}

The negative answer here means that the force is in the direction opposite to the motion, in other words to the left, which is correct. However, the question asks for the magnitude of the force and your answer will be quoted as $\text{170}$ $\text{N}$.