Tug of war

The total mass of the trailers is:
$500 kg + 1000 kg = 1500 kg$
In order for the truck to get the trailers accelerating at 1 meter per second, the truck must exert:
$1500 kg \times 1 \frac{meter}{second} = 1500 Newtons$
In order for trailer A to get trailer B accelerating at 1 meter per second, Trailer A must exert: $500 kg \times 1 \frac{meter}{second} = 500 Newtons$
Finally, dividing the force of the truck by the force of trailer A,
$\frac{1500 Newtons}{500 Newtons} = 3$ , so the final answer is $\boxed{3}.$

We know that according to Newton's second law, $F = ma$

The force exerted by the Truck on Trailor A would be-

$F = m \times a$

$F = 1500 \times 1$

$F = 1500 N$

Now, the force exerted by Trailor A on Trailor B would be-

$F = m \times a$

$F = 500 \times 1$

$F= 500 N.$

Therefore, the required ratio is- $\frac {1500}{500} = 3$

Draw the free body diagram for truck, trailer A, and trailer B.

Let: \overline{Fta} = Contact force from truck to trailer A \overline{Fat} = Contact force from trailer A to truck \overline{Fab} = Contact force from trailer A to trailer B \overline{Fba} = Contact force from trailer B to trailer A \overline{Fm} = Force from the truck machine

Trailer A

\overline{F} = ma \overline{Fta} -\overline{Fba} = 1000 \cdot 1 \overline{Fta} -\overline{Fba} = 1000 ... (1)

Trailer B

\overline{F} = ma \overline{Fab} = 500 \cdot 1 \overline{Fab} = 500 ... (2)

\overline{Fab} = \overline{Fba}

Substitute Eq(2) inti Eq(1)

\overline{Fta} = 1000 + 500 = 1500

\frac {\overline{Fta}}{\overline{Fab}} = \frac {1500}{500} = 3

Balancing of Forces.

common acceleration of the system,a=1 m/s^2
(note:we do not need the value of a in this problem)

force applied by truck on trailer A=(combined mass of trailer A and B) /times (their acceleration).

force applied by trailer A on trailer B=(mass of trailer B) /times (its acceleration).

required ratio=force applied by truck on trailer A / force applied by trailer A on trailer B.

since acceleration of A and B are the same,a cancels out, required ratio = (mass of A + mass of B) / (mass of B) = (1000 + 500) / 500

So,The Given Ratio Of Forces=3.

If T 1 is the force that trailer A exerts on trailer B, then T 1 = 500\times1 = 500 N (N2L on trailer B). If T 2 is the force that the car exerts on trailer A, then T 2-T 1=1000\times1, and so T 2 = 1500 N (N2L on trailer A). Thus the ratio T 2/T 1 is 1500/500=3.

How Newton said, F=m.a. About F on Newton, "m" on kg and "a" on m/s². 1) The truck has 3000kg and accelerate to the right at 1 m/s², so he exerts 3000N force at trailer A. 2) The trailer A has 1000kg and the same acceleration, so the force Trailer A exerts on Trailer B is equal to 1000N. 3) The ratio is truck-trailer A and trailer A-trailer B, so 3000/1000 = 3. So the the ratio of the force the truck exerts on Trailer A to the force Trailer A exerts on Trailer B is equal to 3.

Consider the forces acting on B. The normal force and the weight cancel out. The net force is equal to the force exerted by A.

Consider the forces acting on A and B as a whole. The normal force and the weight cancel out. The net force is equal to the force exerted by the truck.

A and B is thrice as massive as B. A and B has the same acceleration as B. Thus, the net force on A and B is thrice the net force on B.

Therefore, the ratio between the force that the truck exerts on A and the force that A exerts on B is 3.

The truck is only connected with trailer A and cannot exert any direct force on trailer B.

Trailer B has a mass off 500kg and is accelerated by 1m/s², therefore the force on it is F=m*a=500N. This force only can be exerted by trailer A. Therefore trailer B exerts an equal and opposite force on trailer A.

The total force on trailer A has to be 1m/s²*1000kg=1000N to the right. We already know, there is a force of 500N to the left. In consequence the truck has to exert the force 500N+1000N=1500N on trailer A.

After dividing 1500N by 500N, we get the ratio of 3.

Since all the mass(Truck, Trailer-A, Trailer-B) are connected, the following observations should be made clear:

1. The acceleration of all the bodies is the same. i.e., $1m/s^2$ .

2. Since friction is ignored, force on Truck due to its acceleration is equal to the force on Trailer-A.

3.And force on Trailer-A due to its accelerationis equal to force on Trailer-B the .

Here, we don't consider the force on Trailer-A due to the Truck because what force is applied on Trailer-A due to Truck is equal to the force applied on Trailer-B due to Truck. [ Reason: here we are only considering force on Trailer-B due to Trailer-A]

Force the truck exerts on Trailer-A is given by, $F_{AT}=m_{1}a$ where $m_1=3000kg$ and $a = 1m/s^2$ [acceleration is same for all]

$\Rightarrow F_{AT} = 3000N$ [note: $F_{AT}$ is read as force on Trailer A due to Truck]

Similarly, Force Trailer-A exerts on Trailer-B,

$F_{BA} = m_{2}a$ where $m_{2}=1000kg$ . $\Rightarrow F_{BA} = 1000N$

ratio of the force the truck exerts on Trailer A to the force Trailer A exerts on Trailer B is given by, $\frac{F_{AT}}{F_{BA}}= \frac {3000}{1000} = 3$

$\therefore$ The ratio of the force the truck exerts on Trailer A to the force Trailer A exerts on Trailer B is $3$

F=ma, here a = 1 m/s^2 truck pulls 1500 kg so F = 1500 * 1 = 1500 N and trailer pulls 500 kg so f =500 * 1 =500 N The ratio F/f =3

The whole system moves with an acceleration of 1 m/ $s^{2}$ . So, acc.(a)=1 m/ $s^{2}$ .

Now, for the truck, since the truck is carrying its own mass plus the mass of the two trailers, so the force exerted by truck on trailer A = F $_{truck}$ =ma= $(3000+1000+500)\times 1 = 4500 \times 1 = 4500$ N.

Now, for trailer A, it carries its own mass plus mass of trailer B, so the force exerted by trailer A on trailer B = F $_{A}$ = ma = $(1000+500)\times 1 = 1500\times 1 = 1500$ N

Now, Ratio of the forces = $\frac{F_{truck}}{F_{A}} = \frac{4500}{1500} = \boxed{3}$

Force that the truck exerts on trailer A = m(truck) x a = 3000 N

Force that trailer A exerts on trailer B = mt(trailer A ) x a = 1000 N

then ratio is 3000 : 1000 = 3 :1 = 3

Since the acceleration of the body as a whole is the same, we simply equate them. Let f1 and f2 to be forces exerted BY the truck and the trailer respectively and m1, m2 their masses.

Therefore, from the first proposition

f1/m1 = f2/m2

Therefore, f1/f2 = m1/m2 = 3000/1000 = 3.

while F = m * a, where m: is the mass, and a: is the acceleration So for the force the Truck exerts on Trailer A, F1 = 3000 * 1 = 3000 Kg.m/sec^2 and for the force Trailer A exerts on Truck B: F2 = 1000 * 1 = 1000 Kg.m/sec ^ 2

Therefore the ratio = F1/F2 = 3000/1000 = 3.0