A 1200 kg car pulls an 820 kg trailer over a rough road. The
force of friction acting on the trailer is 650 N [backwards].
Calculate the force that the car exerts on the trailer if
(a) the trailer is moving at a constant velocity of
30 km/h [forward]
(b) the trailer is moving at a constant velocity of
60 km/h [forward]
(c) the trailer is moving forward at 60 km/h and starts
accelerating at 1.5 m/s2 [forward]
(d) the trailer is moving forward at 60 km/h and starts
accelerating at 1.2 m/s2 [backwards]
(a) The answer would be 650 N [fwd]. Why? Newton's Third Law States for every action there is an equal and opposite reaction. Also, since Fnet is equal to zero all forces are balanced out. Still don't get it? Draw an FBD stare at it and process all the words I just wrote. Trust me its easier than you make it;)
(b) Still working on it....
b)also 650N [forward] because constant velocity means Fnet is 0.
(a) Well, if the trailer is moving at a constant velocity of 30 km/h forwards, that means the net force acting on it must be zero. So the force that the car exerts on the trailer must be equal in magnitude but opposite in direction to the force of friction acting on the trailer. Therefore, the force that the car exerts on the trailer is 650 N forwards.
(b) If the trailer is moving at a constant velocity of 60 km/h forwards, once again, the net force acting on it must be zero. So the force that the car exerts on the trailer is still 650 N forwards.
(c) Now, if the trailer is moving forward at 60 km/h and starts accelerating at 1.5 m/s² forwards, the net force acting on the trailer is not zero anymore. The force that the car exerts on the trailer must be equal to the sum of the force of friction and the force required to accelerate the trailer. So, in this case, the force that the car exerts on the trailer is 650 N + (820 kg * 1.5 m/s²) forwards.
(d) Lastly, if the trailer is moving forward at 60 km/h and starts accelerating at 1.2 m/s² backwards, the net force acting on the trailer is once again not zero. The force that the car exerts on the trailer must be equal to the sum of the force of friction and the force required to decelerate the trailer. Remember, deceleration is just negative acceleration. So, in this case, the force that the car exerts on the trailer is 650 N + (820 kg * (-1.2 m/s²)) forwards.
Hope that answers your question! Drive safely, and watch out for clowns on the road! 🤡
To calculate the force that the car exerts on the trailer, we need to consider the forces acting on the trailer.
First, let's define the following variables:
m1 = mass of the car = 1200 kg
m2 = mass of the trailer = 820 kg
f_friction = force of friction acting on the trailer = 650 N [backwards]
v = velocity of the trailer = 30 km/h or 60 km/h
a = acceleration of the trailer = 1.5 m/s^2 or 1.2 m/s^2
(a) When the trailer is moving at a constant velocity of 30 km/h [forward]:
Since the velocity is constant, the acceleration is zero. Therefore, the net force on the trailer is zero. This means that the force exerted by the car on the trailer is equal in magnitude but opposite in direction to the force of friction.
So, the force exerted by the car on the trailer is 650 N [forward].
(b) When the trailer is moving at a constant velocity of 60 km/h [forward]:
Similar to the previous case, since the velocity is constant, the acceleration is zero. So, the net force on the trailer is also zero, and the force exerted by the car on the trailer is 650 N [forward].
(c) When the trailer is moving forward at 60 km/h and starts accelerating at 1.5 m/s^2 [forward]:
To calculate the force exerted by the car on the trailer, we need to consider both the force of friction and the force required to accelerate the trailer.
The force required to accelerate the trailer is given by Newton's second law as F = m * a, where m is the mass of the trailer and a is its acceleration.
So, the force required to accelerate the trailer = (820 kg) * (1.5 m/s^2) = 1230 N [forward]
Since the net force on the trailer is the sum of the force of friction and the force required to accelerate it, the force exerted by the car on the trailer is the sum of these forces.
Force exerted by the car on the trailer = 650 N [backwards] + 1230 N [forward]
= 580 N [forward]
(d) When the trailer is moving forward at 60 km/h and starts accelerating at 1.2 m/s^2 [backwards]:
As in the previous case, we need to consider both the force of friction and the force required to accelerate the trailer.
The force required to accelerate the trailer is given by F = m * a, where m is the mass of the trailer and a is its acceleration.
So, the force required to accelerate the trailer = (820 kg) * (-1.2 m/s^2) = -984 N [backwards] (negative sign indicates acceleration in the opposite direction)
Since the net force on the trailer is the sum of the force of friction and the force required to accelerate it, the force exerted by the car on the trailer is the sum of these forces.
Force exerted by the car on the trailer = 650 N [backwards] + (-984 N) [backwards]
= -334 N [backwards] (or 334 N [forwards] with a negative sign indicating opposite direction)