4everaddicted2anime
  • 4everaddicted2anime
If a 990 kg car is on the road and the Ff is 360 N, what is the normal force? I don't understand how to do this.
Physics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Is it a horizontal road or an inclined road?
4everaddicted2anime
  • 4everaddicted2anime
It doesn't say
anonymous
  • anonymous
Than most probably horizontal you have along Y axis W+N=0 therfore N=m.g

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IrishBoy123
  • IrishBoy123
your question doesn't make any sense are you given a coefficient of friction?
anonymous
  • anonymous
It's very odd to include the frictional force in such a problem. Maybe it's just meant to throw you off by including extraneous information. We can find the normal force by drawing a free body diagram and analyzing the forces in the y-direction. |dw:1449178653674:dw| \[\huge \sum F_y = \text{N}-mg=0 \]This equation is equal to zero because it is not accelerating in the y-direction. Therefore:\[\huge \text{N}=mg\]
anonymous
  • anonymous
This is assuming that the system is of a flat, horizontal surface
4everaddicted2anime
  • 4everaddicted2anime
So I would do 360=990x?
anonymous
  • anonymous
Where did you get 360?
anonymous
  • anonymous
Remember that N = mg does not include the frictional force.
anonymous
  • anonymous
You are looking for N, and you are given m. g is the gravitational acceleration constant 9.81 m/s^2
4everaddicted2anime
  • 4everaddicted2anime
Okay. So I do 9.81 × 990 Then I get N equals 9711.9
4everaddicted2anime
  • 4everaddicted2anime
The two answers that have this number are: 9700 N upward 9700 N downward
anonymous
  • anonymous
Yes! There you go! Look at my diagram! We can see that the normal force is the force that the floor exerts on the car (Newton's 3rd Law)
4everaddicted2anime
  • 4everaddicted2anime
I'm only able to see the left portion of the diagram
4everaddicted2anime
  • 4everaddicted2anime
Doesn't a force under an object push it up?
4everaddicted2anime
  • 4everaddicted2anime
Is this the part when I need to use the frictional force to determine whether or not the direction is up or down?
anonymous
  • anonymous
You don't need the frictional force whatsoever, actually. It's extraneous information that is there to possibly confuse you. Dunno why you can only see half of the diagram =/ So according to Newton's 3rd Law, every action has an equal and opposite reaction That is, for our scenario, N=mg. If mg is down, then the reaction, N, must be up!
4everaddicted2anime
  • 4everaddicted2anime
Okay. Thank you for the help
anonymous
  • anonymous
You are very welcome!

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