An 1150-kg car pulls a 450-kg trailer. The car exerts a horizontal force of 3.8 ×10^3 N against

step 1 We have a car pulling a trailer here. So here's my schematic of it. There's a friction force or

An 1150-kg car pulls a 450-kg trailer. The car exerts a...

An $1150-\mathrm{kg}$ car pulls a $450-\mathrm{kg}$ trailer. The car exerts a horizontal force of $3.8 \times 10^{3} \mathrm{N}$ against the ground in order to accelerate. What force does the car exert on the trailer? Assume an effective friction coefficient of 0.15 for the trailer.

Key Concepts

Newton’s Second Law

Newton’s Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This fundamental concept is used to relate the applied force to the acceleration produced, providing the basis for calculating forces in a dynamically accelerating system.

Friction

Friction is the resistive force that opposes motion between two surfaces in contact, and it is calculated as the product of the friction coefficient and the normal force. In problems involving acceleration on a surface, the frictional force must be accounted for because it reduces the effective force available for acceleration.

System of Connected Objects

In systems where objects are connected (such as a car pulling a trailer), both objects share the same acceleration. This concept is essential to understand how forces are transmitted through the connection (for example, the hitch force) and how individual forces, such as friction acting on one part of the system, affect the net force required to accelerate the entire system.

Transcript

00:01 We have a car pulling a trailer here.

00:05 So here's my schematic of it.

00:08 There's a friction force or a drag force on the trailer.

00:13 There's a force exerted on the road by the car to accelerate everything forward.

00:18 We have the weights and the normal forces.

00:21 The car weighs 1150 kilograms.

00:25 The trailer weighs 450 kilograms.

00:28 The force that the car is exerting on the road.

00:32 Road is 3 .8 kiloons and then we're given an effective coefficient of friction which did i write that down here it's 0 .15 so again this is really not really friction between the road and the trailer is just the rolling resistance so they ask us what force does the car exert on the trailer so essentially the force between the car and the trailer here.

01:06 So we draw a free bi diagram.

01:09 So what we can do is the first thing we can do is take the whole, take the system as the car and the trailer together.

01:17 So the external forces acting in the horizontal direction are the force exerted by the car and the friction force on the trailer.

01:25 And then the total mass in that system is the total mass of the car plus the trailer and it's all accelerating at the same rate.

01:33 We can, the friction force, we can get is just the weight of the trailer times the coefficient of friction.

01:42 And now if we look at just the trailer, now t is no longer an internal force.

01:48 It's an external force to this system...

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