Q5-49E - University Physics with Modern Physics

Question

A 1125-kg car and a 2250-kg pickup truck approach a curve on a highway that has a radius of 225 m. 

(a) At what angle should the highway engineer bank this curve so that vehicles traveling at $65.0\text{\hspace{0.17em}mi/h}$ can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car?

(b) As the car and truck round the curve at $65.0\text{\hspace{0.17em}mi/h}$, find the normal force on each one due to the highway surface.

Step-by-Step Solution

Answer

(a) The angle at which highway engineer banks this curve is, ${21.0}^{\circ }$.

(b) The normal force on car and truck are, 11809 N and 23618 N.

Step 1: Identification of given data

The given data can be listed below as,

• The mass of the car is, $m_c=1125\text{\hspace{0.17em}kg}$.
• The radius of the curve is, $R=225\text{\hspace{0.17em}m}$. 
• The mass of the pickup truck is, $m_t=2250\text{\hspace{0.17em}kg}$.
• The speed of the vehicle at the curve is,$v=65.0\text{\hspace{0.17em}mi/h}=29.1\text{\hspace{0.17em}m/s}$ .

Step 2: Significance of normal force

The normal force occurs when two surfaces are in contact, it is the force that one surface exerts on any other surface, known as normal force.

Step 3: Determination of the angle

Part (a)

 

The expression for the vertical component of the normal force in the y-direction can be expressed as,

 $\begin{array}{c}\sum F_y=n\cos \theta +(-mg)=0\\ 0=n\cos \theta =mg\\ n=\frac{mg}{\cos \theta }\end{array}$

Here is the mass of the object,$g$ is the acceleration due to gravity.

 

The expression for the horizontal force in the x-direction can be expressed as,

$\sum F_x=n\sin \theta =m{a}_{rad}$

Here $\frac{mg}{\cos \theta }$ is the centripetal acceleration.

 

Substitute $\frac{mg}{\cos \theta }$ for n, $\frac{v^2}{R}$ for $a_{rad}$ in the above equation.

 $\begin{array}{c}\frac{mg}{\cos \theta }\sin \theta =\frac{m{v}^2}{R}\\ \tan \theta =\frac{v^2}{gR}\\ \theta =ta{n}^{-1}\frac{v^2}{gR}\end{array}$

 

Here $v$ is the speed of the vehicle at the curve, $R$ is the radius of the curve.

 

Substitute $29.1\text{\hspace{0.17em}m/s}$ for $v$, $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$for $g$ , and 225 m for $R$ in the above equation.

$\begin{array}{c}\theta ={\tan }^{-1}\frac{{(29.1\text{\hspace{0.17em}m/s)}}^2}{9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\times 225\text{\hspace{0.17em}m}}\\ \theta ={\tan }^{-1}0.384\\ ={21.0}^{\circ }\end{array}$

Hence, the required angle is,${21.0}^{\circ }$ .

Step 4: determination of normal force on car and truck.

Part (b)

 

The expression for the normal force on a car can be expressed as,

 $n=\frac{m_{c}g}{\cos \theta }$

Here $m_c$ is the mass of the car.

 

Substitute 1125 kg for $m_c$, $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for$g$, and ${21.0}^{\circ }$ for $\theta$ in the above equation.

 $\begin{array}{c}n=\frac{1125\text{\hspace{0.17em}kg }\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}}{\cos {21.0}^{\circ }}\\ =11809\text{\hspace{0.17em}N}\end{array}$

Hence, the required normal force on the car is, 11809 N.

 

The expression for the normal force on the truck can be expressed as,

 $n=\frac{m_{t}g}{\cos \theta }$

Here $m_t$ is the mass of the truck.

 

Substitute 2250 kg for $m_t$ , $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$for $g$ , and ${21.0}^{\circ }$ for $\theta$ in the above equation.

 $\begin{array}{c}n=\frac{2250\text{\hspace{0.17em}kg }\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}}{\cos {21.0}^{\circ }}\\ =23618\text{\hspace{0.17em}N}\end{array}$

Hence, the required normal force on the truck is, 23618 N.