Simple Forces Question

  • Thread starter davedays
  • Start date
  • #1
15
0

Homework Statement


A car of mass 500kg climbs a 15 deg. hill. If the car's engine provides a forward thrust of 8000 N and the drag and friction on the car add up to 800 N calculate the car's rate of acceleration


Homework Equations


F=ma
Normal F= mg x 15 deg



The Attempt at a Solution



Normal force = mg x cos 15

NF = 500 x 10 x 0.9

NF = 4500 N

Backward Force = 800 + mg x sin15

BF = 800 +5000 x 0.2

BF=1800 N

Fwd Force = 8000 N

Resultant force = 8000 N - 1800 = 7200 N


F=ma

7200 = 500 x a
500a=7200
a= 14 ms

I have been killing myself and I still can't find the mistake !

Help me! please !

Thanks,

dave
 

Answers and Replies

  • #2
15
0
Can someone please help me ?
 
  • #3
63
0
What add up to 800 mean?I don't think it means that friction is 800N
 
  • #4
63
0
Maybe that is the error.
 
  • #5
15
0
No mate I just checked twice with the textbook thats exactly ow the question looks like.
 
  • #6
63
0
I don't said that the error was what the textbook said.i said that maybe you have misunderstood it.maybe it doesnot mean that T=800N
 
  • #7
15
0
I think the mistake is somewhere in the resultant force and then there is a follow through :(
 
  • #8
63
0
It says drag nad friction add up to 800
 
  • #9
15
0
so backward force is just 800 ?
 
  • #10
63
0
It cannot because only backward from weight is above 1000.But it may mean that friction is 800 less or more.I donnot know but the mistake must be there
 

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