What is the direction and magnitude of the frictional force?

  • #1
jimmyboykun
39
0

Homework Statement


A 2,200-kg car is moving down a road with a slope (grade) of 14% at a constant speed of 11 m/s.

Homework Equations


The equation that I'm using is ƩF=ma and friction=μN


The Attempt at a Solution


the mass is 2200kg
the acceleration is zero because of the constant speed.
I really cannot think of anyway to solve this solution.
 

Answers and Replies

  • #2
Legaldose
74
6
The frictional force always opposes the motion, it "eats up" the total energy in the system. To do any calculations with the frictional force we need at least the coefficient of friction μ.
 
  • #3
jimmyboykun
39
0
And that the problem, mu is not given. Do I have to find mu in order to move on?
 
  • #4
Legaldose
74
6
Well for this problem you do not need to find mu. Here are some steps to get you started:

1. Draw a free body diagram on the car.
2. Utilize Newtons Second Law in the x and y directions respectively
3. Algebraic Wizardry, manipulate the equations until you can find the frictional force using values you already know.
 
  • #5
jimmyboykun
39
0
ok, but what about the 11m/s, does that velocity come into play?
 
  • #6
NihalSh
199
15
ok, but what about the 11m/s, does that velocity come into play?

your question mentions that velocity is constant, and when velocity is constant the net force is always zero. so, here friction is equal to the component of gravitational force along the slope.

P.S. - like Legaldose already mentioned drawing free body diagram always clears out doubt.
 
  • #7
jimmyboykun
39
0
ok so I solved the problem, and this is what I did: m*g*sin(7.96) the 7.96 is the reverse tangent.

so 2,200kg*9.81m/s^2*sin(7.96)= 2,989N

would that be considered my magnitude and direction?
 
  • #8
rcgldr
Homework Helper
8,791
579
ok so I solved the problem, and this is what I did: m*g*sin(7.96) the 7.96 is the reverse tangent.
Not a big deal, but arctan(.14) ~= 7.97.
 
  • #9
jimmyboykun
39
0
thanks, but what about the solution that I solved 2,989N is that solution the magnitude and direction, or if there is there still more?
 
  • #10
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
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7,954
The frictional force always opposes the motion,
That formulation can confuse people when thinking about e.g. a car being driven up a slope. It leads them to think that the friction between road and tyre would act down the slope. Better always to write that friction opposes relative motion of the surfaces in contact.
 
  • #11
NihalSh
199
15
thanks, but what about the solution that I solved 2,989N is that solution the magnitude and direction, or if there is there still more?

it's the magnitude you have calculated. For direction refer to haruspex's post above.
 

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