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<h2>Example: Newton's Second Law: Truck and Trailer</h2>
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<h3>Question</h3>
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<p>A $\text{2 000}$ $\text{kg}$ truck pulls a $\text{500}$ $\text{kg}$ trailer with a constant acceleration. The engine of the truck produces a thrust of $\text{10 000}$ $\text{N}$. Ignore the effect of friction. Calculate the:</p>
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<p>acceleration of the truck; and</p>
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<p>tension in the tow bar T between the truck and the trailer, if the tow bar makes an angle of $\text{25}$$\text{°}$ with the horizontal.</p>
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<div class="wp-caption alignnone" style="width: 405px"><img alt="61af3c0c1fd4bdcbfb06aaa930371b4e.png" height="223" src="https://data.ulearngo.com/assets/7a4713a1-0b1d-4945-b7ff-a4ee09c7bb17" width="405"/><p class="wp-caption-text">Truck pulling a trailer.</p></div>
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<h3>Step 1: Draw a force diagram</h3>
<p>Draw a force diagram indicating all the forces on the system as a whole:</p>
<div class="wp-caption alignnone" style="width: 478px"><img alt="18873dbb7b4177949581ebfd1a11cf1a.png" height="200" src="https://data.ulearngo.com/assets/9020c1ae-6128-4c9b-a6ae-47da47112260" width="478"/><p class="wp-caption-text">Free body diagrams for truck pulling a trailer.</p></div>
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<h3>Step 2: Apply Newton's second law of motion</h3>
<p>We choose the positive $x$-direction to be the positive direction. We only need to consider the horizontal forces. Using only the horizontal forces means that we first need to note that the tension acts at an angle to the horizontal and we need to use the horizontal component of the tension in our calculations.</p>
<img alt="f19f1f4d3036f2290bb6f5d808b33830.png" src="https://data.ulearngo.com/assets/52d0b3b5-a125-480d-bbac-a33fec75545c"/>
<p>The horizontal component has a magnitude of $T\cos(\text{25}\text{°})$.</p>
<p>In the absence of friction, the only force that causes the system to accelerate is the thrust of the engine. If we now apply Newton's second law of motion to the truck we have: \begin{align*} \vec{F}_{Rtruck} &amp;= m_{truck}\vec{a}\ \text{(we use signs to indicate direction)} \\ {F}_{engine} - T\cos(\text{25}\text{°}) &amp;= (\text{2 000})a \\ (\text{10 000}) - T\cos(\text{25}\text{°}) &amp;= (\text{2 000})a \\ a &amp; = \frac{(\text{10 000}) - T\cos(\text{25}\text{°})}{(\text{2 000})} \end{align*}</p>
<p>We now apply the same principle to the trailer (remember that the direction of the tension will be opposite to the case of the truck): \begin{align*} \vec{F}_{Rtrailer} &amp;= m_{trailer}\vec{a}\ \text{(we use signs to indicate direction)} \\ T\cos(\text{25}\text{°}) &amp;= (\text{500})a \\ a &amp; = \frac{T\cos(\text{25}\text{°})}{(\text{500})} \end{align*}</p>
<p>We now have two equations and two unknowns so we can solve simultaneously. We subtract the second equation from the first to get: \begin{align*} (a) - (a) &amp;=(\frac{(\text{10 000}) - T\cos(\text{25}\text{°})}{(\text{2 000})}) - (\frac{T\cos(\text{25}\text{°})}{(\text{500})}) \\ 0 &amp; = (\frac{(\text{10 000}) - T\cos(\text{25}\text{°})}{(\text{2 000})}) - (\frac{T\cos(\text{25}\text{°})}{(\text{500})})\\ &amp; \text{(multiply through by 2 000)} \\ 0 &amp; = (\text{10 000}) - T\cos(\text{25}\text{°}) - 4T\cos(\text{25}\text{°}) \\ \text{5}T\cos(\text{25}\text{°}) &amp; = (\text{10 000}) \\ T &amp; = \frac{(\text{10 000})}{\text{5}\cos(\text{25}\text{°})} \\ T &amp; = \text{2 206,76}\text{ N} \end{align*}</p>
<p>Now substitute this result back into the second equation to solve for the magnitude of $a$ \begin{align*} a &amp; = \frac{T\cos(\text{25}\text{°})}{(\text{500})}\\ &amp; = \frac{(\text{2 206,76})\cos(\text{25}\text{°})}{(\text{500})}\\ &amp; = \text{4,00}\text{ m·s$^{-2}$} \end{align*}</p>
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Truck and Trailer Example

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Truck and Trailer Example

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