Question

A 1230kg car pushes a 2160kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4590N . Take the friction between the truck's tires and the ground to be 470N .

What is the magnitude F_ConT of the force that the car exerts on the truck? Use the coordinate system shown in Part B, where upward is the positive y direction and to the right, the direction F_ConT points, is the positive x direction.

Answer 1

The magnitude of the force that the car exerts on the truck ( F_C on T ) is 3095 N

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

This problem is about Newton's Law of Motion.

Given:

mass of car = $m_c$ = 1230 kg

mass of truck = $m_t$ = 2160 kg

force of car = $f_c$ = 4590 N

friction of truck = $f_t$ = 470

Unknown:

$F_{C ~ on ~ T}=?$

Solution:

Car :

$\Sigma F = ma$

$f_c - F_{T ~ on ~ C} = m_c a$

$4590 - F_{T ~ on ~ C} = 1230 a$

$\boxed {a = ( 4590 - F_{T ~ on ~ C} ) \div 1230}$ → Equation 1

Truck :

$\Sigma F = ma$

$F_{C ~ on ~ T} - f_t = m_t a$

$F_{C ~ on ~ T} - 470 = 2160 a$

$F_{C ~ on ~ T} - 470 = 2160 ( 4590 - F_{T ~ on ~ C} ) \div 1230$

In accordance with Newton's 3rd law, then $F_{C ~ on ~ T} = F_{T ~ on ~ C}$

$1230 ( F_{C ~ on ~ T} - 470 ) = 2160 ( 4590 - F_{C ~ on ~ T} )$

$(1230 + 2160)F_{C ~ on ~ T} = (2160 \times 4590) + (1230 \times 470)$

$3390F_{C ~ on ~ T} = 10492500$

$F_{C ~ on ~ T} = 10492500 \div 3390$

$\large {\boxed {F_{C ~ on ~ T} \approx 3095 ~ N} }$

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Mug , Friction , Coefficient , Static

Answer 2

The magnitude of the force applied by car on the truck is $\fbox{\begin\\3105.2\,{\text{N}}\end{minispace}}$.

Further Explanation:

The car applies a force on the ground due to which the truck moves in the positive x direction. The friction force between the tires of the truck and ground oppose the external force applied by the car. Therefore, the net force on the car truck system is the difference between the force applied by the car on the ground and the friction force on the tires of the truck and it is acting along the direction of motion of the car.

Given:

The mass of the car is $1230\,{\text{kg}}$.

The mass of the truck is $2160\,{\text{kg}}$.

The force with which the car pushes the ground is $4590\,{\text{N}}$.

The friction force between the tires of the truck and ground is $470\,{\text{N}}$.

Concept:

The net force on the car truck system is defined as the product of net mass of the system i.e., sum of mass of car and mass of truck and the acceleration with which the system is moving.

The net force on the car truck system is:

${F_{net}} = \left( {{m_C} + {m_T}} \right)a$

Rearrange the above expression.

$a=\dfrac{{{F_{net}}}}{{\left( {{m_C} + {m_T}} \right)}}$                                             …… (1)

Here, ${F_{net}}$ is the net force on the car truck system, ${m_C}$ is the mass of the car, ${m_T}$ is the mass of the truck and $a$ is the acceleration with which the system is moving.

The net force on the car truck system is:

${F_{net}} = {F_C} - f$                                                                               …… (2)

Here, ${F_C}$ is the force with which the car pushes the ground and $f$ is the friction force between the truck tires and ground.

The force applied by the car on the truck is:

${F_{C\,on\,T}} - f = {m_T}\text{ a}$  

Rearrange the above expression.

${F_{C\,on\,T}} = {m_T}\text{ a}+f$                                                                                …… (3)

Here, ${F_{C\,on\,T}}$ is the force applied by car on truck.

Substitute the values of ${F_C}$ and f in equation (2).

$\begin{aligned}{F_{net}}&=4590\,{\text{N}} - 470\,{\text{N}}\\&=4120\,{\text{N}}\\\end{aligned}$

Substitute the value of ${F_{net}}$, ${m_C}$ and ${m_T}$ in equation (1).

$\begin{aligned}a&=\frac{{4120\,{\text{N}}}}{{\left( {1230\,{\text{kg}} + 2160\,{\text{kg}}} \right)}} \\&=1.22\,{{\text{m}}\mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}} \\\end{aligned}$

Substitute the values of ${m_T}$, $a$ and $f$ in equation (3).

$\begin{aligned}{F_{C\,on\,T}}&=\left( {2160\,{\text{kg}}} \right)\left( {1.22\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}} \right)+\left( {470\,{\text{N}}} \right)\\&=3105.2\,{\text{N}} \\ \end{aligned}$

Thus, the magnitude of the force applied by car on the truck is $\fbox{\begin\\3105.2\,{\text{N}}\end{minispace}}$.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Motion, force, friction force, car, truck, force applied by car on truck, force applied by truck on car, acceleration, 3105.2 N, 3105 N, 3105.2 newton, 3105 newton, 3105.2 Newton, 3105 Newton.