What is the magnitude of the force of the car on the truck?

AI Thread Summary
A 1100 kg car exerts a force of 4500 N to push a 1700 kg truck with a dead battery. The acceleration of the system is calculated to be 1.607 m/s² using the equation F=ma. The force of the truck on the car is derived from the net force acting on the car, leading to a calculated value of 2732.3 N. Participants discuss the importance of correctly identifying forces, particularly the forward force exerted by the car. The conversation emphasizes understanding the dynamics of the system and the relationships between the forces involved.
aligass2004
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Homework Statement


A 1100 kg car pushes a 1700 kg 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 4500N. a.) What is the magnitude of the force of the car on the truck? b.) What is the magnitude of the force of the truck on the car?


Homework Equations


F=ma

The Attempt at a Solution


I tried drawing a free body diagram for this, but it didn't work out well.
 
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First do the freebody diagram of the truck+car system. That gives you acceleration...

Then do the freebody diagram of the car alone. that'll give the force of the truck on the car.
 
I got the acceleration to be 1.607 m/s^2. I tried using F=ma, but it didn't work.
 
aligass2004 said:
I got the acceleration to be 1.607 m/s^2. I tried using F=ma, but it didn't work.

The acceleration is correct. What equation did you get when you used F=ma for the car?
 
The number I got was 1767.7N
 
aligass2004 said:
The number I got was 1767.7N

That's not right. What are the forces acting on the car?
 
The forces acting on the car are weight, the normal force, and the force of the truck.
 
aligass2004 said:
The forces acting on the car are weight, the normal force, and the force of the truck.

You're missing one force. The main one.
 
I don't understand.
 
  • #10
aligass2004 said:
I don't understand.

The 4500N.
 
  • #11
Ok, so the normal force equal 4500N as well. So how do I go about finding the force of the truck on the car with all of that information?
 
  • #12
aligass2004 said:
Ok, so the normal force equal 4500N as well. So how do I go about finding the force of the truck on the car with all of that information?

It's not the normal force... It is a forward force exerted at the back of the car... (it's the same force you used for the truck+car part).

Write out the F = ma equation for the car in the horizontal direction.
 
  • #13
I am still confused.
 
  • #14
aligass2004 said:
I am still confused.

There's a 4500N forward force at the back of the car... there's a backward force from the truck.

\Sigma{F} = ma

4500 - F_{truck} = m_{car}a
 
  • #15
I got the answer to be 2732.3, but I still understand the forward force at the back of the car thing.
 
  • #16
aligass2004 said:
I got the answer to be 2732.3, but I still understand the forward force at the back of the car thing.

How did you do the first part?
 
  • #17
I found the acceleration.
 
  • #18
aligass2004 said:
I found the acceleration.

Yes, how did you find the acceleration? You used the 4500N force... how did you use it... how did you know how to use it?
 
  • #19
I used F=ma and manipulated the equation around to a=F/m. I don't know why I knew to use it though.
 
  • #20
aligass2004 said:
I used F=ma and manipulated the equation around to a=F/m. I don't know why I knew to use it though.

Draw a picture... the 4500N force is pushing the car forward... and the car is pushing the truck forward...
 
  • #21
OHHHHHHHHHHHHHHHHH! I was putting the 4500N as down. I see now. Wow, I feel completely stupid. Sorry about that.
 
  • #22
aligass2004 said:
OHHHHHHHHHHHHHHHHH! I was putting the 4500N as down. I see now. Wow, I feel completely stupid. Sorry about that.

lol! That's ok. :smile:
 
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