Finding frictional force towing a car up a hill

  • #1

Homework Statement



A tow truck driver hooks up a car to tow it up the hill. The car weighs 1500kg. the cable makes an angle of28 degrees with ground. the cables tension is 1900 N . the static coefficitient is 0.5 and the acceleration is at 6/m s^2 . what is the frictional force on the car

Homework Equations



ƩF=ma


The Attempt at a Solution



using my diagram and summing the forces in the xdirection i get ƩF= Tension + Fgravity- Force of Friction= ma

1900sin28-9.8sin28-0.5N=1900 X 6 ...then I solve for N?
 

Answers and Replies

  • #2
cepheid
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I'm assuming that you mean that the cable is parallel to the incline, and that the incline makes an angle of 28 deg with the *horizontal?* If not, can you clarify please?

Also, are you taking your x direction to be the direction parallel to the incline? If so, why does the cable tension (1900 N) have a sine factor multiplying it in your equations?

Thanks,

cepheid
 
  • #3
I'm assuming that you mean that the cable is parallel to the incline, and that the incline makes an angle of 28 deg with the *horizontal?* If not, can you clarify please?
Yes,
the cable is parallel to the incline.
Also, are you taking your x direction to be the direction parallel to the incline? If so, why does the cable tension (1900 N) have a sine factor multiplying it in your equations?

Thanks,

cepheid
yes i made mistake tension should be 1900cos28 i.e.
1900cos28-9.8sin28-0.5N=1900 X 6 ...then I solve for N?
so is that pretty much the process to solve for frictional force?
 
  • #4
PhanthomJay
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You still need to clarify and answer cepheid's questions. Check the numbers given on the problem. The values you list make no sense. If the 1500 kg car was accelerating at 6m/s/s , a net force of 9000 N is required to accelerate it at that racecar rate .
 
  • #5
I'm assuming that you mean that the cable is parallel to the incline, and that the incline makes an angle of 28 deg with the *horizontal?* If not, can you clarify please?

Also, are you taking your x direction to be the direction parallel to the incline? If so, why does the cable tension (1900 N) have a sine factor multiplying it in your equations?

Thanks,

cepheid


Yes, the incline is 28 with respect to the horizontal so 28 degrees from 0 degrees. I put my x direction to be parallel to the incline so my T vector is at 28, my N vector is at 90 degrees , frictional force is at 180. and Fg is at 270.

Therefore, to find the frictional force do I use Ff= [itex]\mu[/itex] times N= mgcostheta , therefore Ff = [itex]\mu[/itex]*1500*-9.8 cos 28 therefore my force of friction is 8490N ?
 
Last edited:
  • #6
cepheid
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Yes, the incline is 28 with respect to the horizontal so 28 degrees from 0 degrees. I put my x direction to be parallel to the incline so my T vector is at 28, my N vector is at 90 degrees , frictional force is at 180. and Fg is at 270.

No, the normal force is not at 90 degrees (straight vertical). The word normal means "perpendicular." The normal force is a contact force that acts perpendicular to the two surfaces that are in contact. So, in this case, the normal force is perpendicular to the incline.

Same with the frictional force. The frictional force is parallel to the incline, not at 180 degrees.

Therefore, to find the frictional force do I use Ff= [itex]\mu[/itex] times N= mgcostheta , therefore Ff = [itex]\mu[/itex]*1500*-9.8 cos 28 therefore my force of friction is 8490N ?

This is the maximum frictional force that the surface of plane could provide to prevent the car from moving, but that's not relevant here, because the car is moving. You need to use the force balance (Newton's second law) to determine the frictonal force from the tension and net force.

BUT, as PhantomJay said above, the numbers in your problem make no sense, because a force of 9000 N is required to give the car that huge acceleration of 6 m/s^2, and the tensiion is way less than that. You need to double check your numbers.
 

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