Chapter 4, Problem 69GP

To determine

The force exerted by the car on the trailer.

Answer to Problem 69GP

The force exerted by the car:

  $1.5\times {10}^3\text{N}$

Explanation of Solution

Given Data:

  $\begin{array}{l}\text{mass of Car}\left(M_{car}\right)=1150\text{kg}\\ \text{mass}\text{of}\text{trailer}\left(m_{trailer}\right)=450\text{kg}\end{array}$

The horizontal force exerted by the car:

  $3.8\times {10}^3\text{N}$ against the ground.

The effective friction coefficient =0.15

Calculation:

The free body diagram for the given car and trailer is drawn below:

  

Where,

  $\begin{array}{l}{F}_{CG}=\text{Force on car due to ground}\\ F_{TC}=\text{Force on trailer due to car}\\ F_{CT}=\text{Force on car due to trailer}\end{array}$

Using the Newton’s third law:

  $\left|F_{TC}\right|=\left|F_{CT}\right|$

Referring to the free body diagram, apply the Newton’s second law:

  $\begin{array}{l}x-\text{component}\\ F-F_T=M_{C}a\mathrm{............}\left(1\right)\\ y-\text{component}\\ F_{Ncar}-M_{Cg}=0\mathrm{............}\left(2\right)\\ \text{For trailer:}\\ x-\text{component}\\ F-F_{frT}=m_{T}a\mathrm{............}\left(3\right)\\ y-\text{component}\\ F_{N_T}=m_{T}g\mathrm{............}\left(4\right)\end{array}$

Since, the normal force will be equal to the weight, hence the frictional force:

  $\begin{array}{l}{F}_{fr}={\mu }_k{F}_{NT}\\ ={\mu }_k{m}_{T}g\\ =0.15\left(450\text{kg}\right)\left(9.8{\text{m/s}}^2\right)\\ =660\text{N}\end{array}$

Now, adding equation 1 and 3:

  $\begin{array}{l}F-F_{frT}=a\left(M_C+m_T\right)\\ a=\frac{F-F_{frT}}{M_C+m_T}\\ =1.96{\text{m/s}}^2\end{array}$

Considering the free body diagram for the trailer alone.

Substitute the known values into the above equation:

  $\begin{array}{c}{F}_T-{\mu }_s{F}_{NT}=m_{T}a\\ F_T=m_{T}a+{\mu }_s{m}_{T}g\\ =m_T\left(a+{\mu }_{s}g\right)\\ =450\left(1.96+0.15\left(9.8\right)\right)\\ =1.5\times {10}^3\text{N}\end{array}$