• Support PF! Buy your school textbooks, materials and every day products Here!

Help with F=ma

  • Thread starter xdeanna
  • Start date
  • #1
24
0
Each of the four wheels of a car pushes on the road with a force of 4.0 x 10³ N [down]. The driving force on the car is 8.0 x 10³ N [W]. The frictional resistance on the car is 6.0 x 10³ N [E]. Calculate the following:
(a) mass of the car
(b) net force on the car
(c) car's acceleration

For Fnet, would u also add 4(4.0 x 10³ N [up]) to the rest of vectors to cancel out the force all the tires have on the road? Or:
8.0 x 10³ N [W] + 6.0 x 10³ N [E]= 2.0 x 10³ N [W]

2.0 x 10³ N [W] + 4(4.0 x 10³ N [down])= Fnet
?!

I just did all that on my phone.. I'm desperate for some help :(
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31


Each of the four wheels of a car pushes on the road with a force of 4.0 x 10³ N [down]. The driving force on the car is 8.0 x 10³ N [W]. The frictional resistance on the car is 6.0 x 10³ N [E]. Calculate the following:
(a) mass of the car
(b) net force on the car
(c) car's acceleration

For Fnet, would u also add 4(4.0 x 10³ N [up]) to the rest of vectors to cancel out the force all the tires have on the road? Or:
8.0 x 10³ N [W] + 6.0 x 10³ N [E]= 2.0 x 10³ N [W]

2.0 x 10³ N [W] + 4(4.0 x 10³ N [down])= Fnet
?!

I just did all that on my phone.. I'm desperate for some help :(
I am led to believe that the entire mass of the car is supported by the wheels alone.

So yes the entire weight is 4(4x103) N, so the mass of the car is ?
 
  • #3
24
0


I am led to believe that the entire mass of the car is supported by the wheels alone.

So yes the entire weight is 4(4x103) N, so the mass of the car is ?
m= 1.6 x 104 N

but what about Fnet
 
  • #4
rock.freak667
Homework Helper
6,230
31


m= 1.6 x 104 N

but what about Fnet
There is a driving force and a resistance to the driving force, so what should be the net force between these two?
 
  • #5
24
0


There is a driving force and a resistance to the driving force, so what should be the net force between these two?
so i don't have to worry about the force of the tires on the road?

Fnet=8.0 x 10³ N [W] + 6.0 x 10³ N [E]
Fnet= 2.0 x 10³ N [W]
 
  • #6
rock.freak667
Homework Helper
6,230
31


so i don't have to worry about the force of the tires on the road?

Fnet=8.0 x 10³ N [W] - 6.0 x 10³ N [E]
Fnet= 2.0 x 10³ N [W]
you mean negative here


But yes that should be correct.
 

Related Threads for: Help with F=ma

  • Last Post
Replies
4
Views
2K
  • Last Post
3
Replies
59
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
9K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
Top