6.1 Solving problems with newton’s laws  (Page 9/12)

What force acts on a model helicopter?

A 1.50-kg model helicopter has a velocity of $5.00\widehat{j}\text{m/s}$ at $t=0.$ It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of $\left(6.00\widehat{i}+12.00\widehat{j}\right)\text{m/s}\text{.}$ What is the magnitude of the resultant force acting on the helicopter during this time interval?

Strategy

We can easily set up a coordinate system in which the x -axis $(\widehat{i}$ direction) is horizontal, and the y -axis $(\widehat{j}$ direction) is vertical. We know that $\text{Δ}t=2.00s$ and $(6.00\widehat{i}+12.00\widehat{j}\text{m/s})-(5.00\widehat{j}\text{m/s}).$ From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

Solution

We have

The magnitude of the force is now easily found:

Significance

The original problem was stated in terms of $\widehat{i}-\widehat{j}$ vector components, so we used vector methods. Compare this example with the previous example.

Got questions? Get instant answers now!

Baggage tractor

[link] (a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by $F=(820.0t)\text{N,}$ find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

Strategy

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force $\overrightarrow{T},$ which is our objective.

Solution

1. ${\displaystyle \sum F_x}=m_{\text{system}}{a}_x$ and ${\displaystyle \sum F_x}=820.0t,$ so
   
   $\begin{array}{ccc}\hfill 820.0t& =\hfill & (650.0+250.0+150.0)a\hfill \\ \hfill a& =\hfill & 0.7809t.\hfill \end{array}$
   
   Since acceleration is a function of time, we can determine the velocity of the tractor by using $a=\frac{dv}{dt}$ with the initial condition that $v_0=0$ at $t=0.$ We integrate from $t=0$ to $t=3\text{:}$
   
   $dv=adt,{\displaystyle {\int }_0^{3}dv=}{\displaystyle {\int }_0^{3.00}adt}={\displaystyle {\int }_0^{3.00}0.7809tdt},v=0.3905{t^2]}_0^{3.00}=3.51\text{m/s}\text{.}$
2. Refer to the free-body diagram in [link] (b).
   
   $\begin{array}{ccc}\hfill {\displaystyle \sum F_x}& =\hfill & m_{\text{tractor}}{a}_x\hfill \\ \hfill 820.0t-T& =\hfill & m_{\text{tractor}}(0.7805)t\hfill \\ \hfill (820.0)(3.00)-T& =\hfill & (650.0)(0.7805)(3.00)\hfill \\ \hfill T& =\hfill & 938\text{N}.\hfill \end{array}$

Significance

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving [link] (a), whereas only the mass of the truck (since it supplied the force) was of use in [link] (b).

Got questions? Get instant answers now!