Question

A 960 kg car is pulling a 310 kg trailer. Together the car and trailer move forward with an acceleration of 2.00 m/s2. Ignore any frictional force of air drag on the car and all frictional forces on the trailer.
(a) Determine the net force on the car.
magnitude

N
direction


(b) Determine the net force on the trailer.
magnitude

N
direction


(c) Determine the force exerted by the trailer on the car.
magnitude

N
direction


(d) Determine the resultant force exerted by the car on the road. (Assume that the forward direction is along the +x-direction.)
magnitude

Answer 1

Answer:

(a) 2540 N

(b) 620 N

Explanation:

mass of car, M = 960 kg

mass of trailer, m = 310 kg

acceleration, a = 2 m/s^2

Let the net force on the car is F and the net force on the trailer is T.

(a) According to the free body diagram, use Newton's second law

F - T = Ma .... (1)

T = ma   .... (2)

Adding both the equations

F = (M+m)a

F = (960 + 310) x 2

F = 2540 N

Thus, the force on the car is 2540 N.

(b) By the equation (2), we get

T = 310 x 2 = 620 N

Thus, the force on trailer is 620 N.

Answer 2

Answer:

a) $F_c=1920\,N$

b) $F_c=620\,N$

c) $T=620\,N$

d) $F_R=9497.39\,N$          

    $\theta=81.133^{\circ}$ from the horizontal.

Explanation:

Given:

mass of the car, $m_c=960\,kg$

mass of trailer,  $m_t=310\,kg$

acceleration of the system,  $a=2\,m.s^{-2}$

(a)

Net force on the car will be in the direction of acceleration given by:

$F_c=m_c.a$

$F_c=960\times 2$

$F_c=1920\,N$

(b)

$F_t=m_t.a$

$F_t=310\times 2$

$F_t=620\,N$

(c)

Since the car and the trolley are linked together, the link will face a tension force and this will be exerted on the car by the Newton's third law of motion which can be given as:

$T=2\times 310$

$T=620\,N$

(d)

Net vertical force acting on the road due to car:

$F_v=m_c.g$

$F_v=960\times 9.8$

$F_v=9408\,N$

Net vertical force acting on the road:

$F_h=(960-310)\times 2$

$F_h=1300\,N$

Now, the net resultant force:

$F_R=\sqrt{(1300)^2+(9408)^2}$

$F_R=9497.39\,N$

Angle f the force from horizontal:

$tan\theta=\frac{9408}{1300}$

$\theta=81.133^{\circ}$ from the horizontal.