# Simple Forces Question

1. Nov 9, 2009

### davedays

1. The problem statement, all variables and given/known data
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

2. Relevant equations
F=ma
Normal F= mg x 15 deg

3. 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 !

Thanks,

dave

2. Nov 9, 2009

3. Nov 9, 2009

### Tzim

What add up to 800 mean?I don't think it means that friction is 800N

4. Nov 9, 2009

### Tzim

Maybe that is the error.

5. Nov 9, 2009

### davedays

No mate I just checked twice with the textbook thats exactly ow the question looks like.

6. Nov 9, 2009

### Tzim

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. Nov 9, 2009

### davedays

I think the mistake is somewhere in the resultant force and then there is a follow through :(

8. Nov 9, 2009

9. Nov 9, 2009

### davedays

so backward force is just 800 ?

10. Nov 9, 2009

### Tzim

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