## Example: Newton’s Second Law: Box on a Surface

Contents

### 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: Evaluate what is given

To make things easier lets give the two crates labels, let us call the \(\text{10}\) \(\text{kg}\) crate number 2 and the \(\text{15}\) \(\text{kg}\) crate number 1.

We have two crates that have overall has an acceleration that is given. The fact that the crates are tied together with a rope means that they will both have the same acceleration. They will also both feel the same force due to the tension in the rope.

We are told that there is friction but we are only given the relationship between the total frictional force both crates experience and the fraction each one experiences. The total friction, \(\vec{F}_{fT}\) will be the sum of the friction on crate 1, \(\vec{F}_{f1}\), and the friction on crate 2, \(\vec{F}_{f2}\). We are told that \(\vec{F}_{f1}=\frac{\text{2}}{\text{3}}\vec{F}_{fT}\) and \(\vec{F}_{f2}=\frac{\text{1}}{\text{3}}\vec{F}_{fT}\). We know the blocks are accelerating to the right and we know that friction will be in the direction opposite to the direction of motion and parallel to the surface.

### Step 2: Draw force diagrams

The diagram for crate 1 will be:

The diagram for crate 1 (indicated by blue dotted lines) will be:

Where:

- \(\vec{F}_{g1}\) is the force due to gravity on the first crate
- \(\vec{N}_{1}\) is the normal force from the surface on the first crate
- \(\vec{T}\) is the force of tension in the rope
- \(\vec{F}_{applied}\) is the external force being applied to the crate
- \(\vec{F}_{f1}\) is the force of friction on the first crate

The diagram for crate 2 (indicated by orange dashed lines) will be:

Where:

- \(\vec{F}_{g2}\) is the force due to gravity on the second crate
- \(\vec{N}_{2}\) is the normal force from the surface on the second crate
- \(\vec{T}\) is the force of tension in the rope
- \(\vec{F}_{f2}\) is the force of friction on the second crate

### Step 3: Apply Newton’s second law of motion

The problem tells us that the crates are accelerating along the \(x\)-direction which means that the forces in the \(y\)-direction do not result in a net force. We can treat the different directions separately so we only need to consider the \(x\)-direction.

We are working with one dimension and can choose a sign convention to indicate the direction of the vectors. We choose vectors to the right (or in the positive \(x\)-direction) to be positive.

We can now apply Newton’s second law of motion to the first crate because we know the acceleration and we know all the forces acting on the crate. Using positive to indicate a force to the right we know that \({F}_{res1} = F_{applied}-{F}_{f1}-T\)\begin{align*} \vec{F}_{res1} &=m_1\vec{a} \\ F_{applied}-{F}_{f1}-T &=m_1a \\ F_{applied}-\frac{\text{2}}{\text{3}}{F}_{fT}-T &=m_1a \\ (500)-\frac{\text{2}}{\text{3}}{F}_{fT}-T &=(\text{15})(2) \\ -T &=(\text{15})(2) -(500)+\frac{\text{2}}{\text{3}}{F}_{fT}\end{align*}

Now apply Newton’s second law of motion to the second crate because we know the acceleration and we know all the forces acting on the crate. We know that \({F}_{res2} = T-{F}_{f2}\). **Note that tension is in the opposite direction.**\begin{align*} \vec{F}_{res2} &=m_2\vec{a} \\ T-{F}_{f2} &=m_2a \\ T – \frac{\text{1}}{\text{3}}{F}_{fT} &=m_2a \\ T &=(\text{10})(2) +\frac{\text{1}}{\text{3}}{F}_{fT}\end{align*}

### Step 4: Solve simultaneously

We have used Newton’s second law of motion to create two equations with two unknowns, this means we can solve simultaneously. We solved for \(T\) in the equations above but one carries a negative sign so if we add the two equations we will subtract out the value of the tension allowing us to solve for \({F}_{fT}\): \begin{align*} (T) + (-T) &= ((\text{10})(2) +\frac{\text{1}}{\text{3}}{F}_{fT}) + ((\text{15})(2) -(500)+\frac{\text{2}}{\text{3}}{F}_{fT}) \\ 0 & = \text{20} + \text{30} – \text{500} + \frac{\text{1}}{\text{3}}{F}_{fT} + \frac{\text{2}}{\text{3}}{F}_{fT} \\ 0 & = -\text{450} + {F}_{fT} \\ {F}_{fT} & = \text{450}\text{ N} \end{align*}

We can substitute the magnitude of \({F}_{fT}\) into the equation for crate 2 to determine the magnitude of the tension: \begin{align*} T &=(\text{10})(2) +\frac{\text{1}}{\text{3}}{F}_{fT} \\ T &=(\text{10})(2) +\frac{\text{1}}{\text{3}}(\text{450}) \\ T &= \text{20} +\text{150} \\ T &= \text{170}\text{ N} \end{align*}

### Step 5: Quote final answers

The total force due to friction is \(\text{450}\) \(\text{N}\) to the left. The magnitude of the force of tension is \(\text{170}\) \(\text{N}\).