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Normal Force Question

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data

    I've been having a little bit of trouble answering this question:

    "An 8.0 kg wooden sled is pulled over the snow by means of a rope that makes an angle of 40° with the horizontal. If the rope has a tension of 70.0 N, what is the normal force acting on the sled?"

    2. Relevant equations

    w = mg

    , gives the weight of an object

    FN = mg cos(θ)

    , gives the normal force acting on an object on an inclined surface. But since it hasn't been mentioned in my textbook yet, I'm thinking that the question has to be answered using vector components only.

    X = R cos(θ)

    , and:

    Y = R sin(θ)

    , gives the horizontal and vertical components of a vector respectively.

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

    Unfortunately, I don't have a picture of my diagram, but what I first drew is a triangle with a 40° angle corner, representing the hill.

    I drew a vector pointing straight down, starting from the centre of the sled on the hill, marked "78.48 N", which is the weight of the sled.

    I drew another vector pointing upwards, perpendicular to the hill surface, starting from the centre of the sled, representing the normal force which needs to be solved.

    On the side of the sled, I drew a vector, parallel to the hypotenuse of the hill, marked "70 N", which is the tension on the rope. I formed a triangle using that side and, by checking, I found that the corner angle of this triangle was 50°.

    Lastly, above and beneath the diagram of the hill, I joined the weight vector, which acted as the hypotenuse, with the normal force vector, which acted as the adjacent side, forming a triangle.

    3. The attempt at a solution

    First, I converted the mass of the object into its weight:

    w = mg

    w = 8.0 * - 9.81

    w = -78.48 N

    This is where the problem began.

    Now that I had a vector triangle with the hypotenuse calculated to be -78.48 N and a corner angle of 40°, I used the normal force formula to find the adjacent side (the normal force excluding the tension of the rope). And since the calculated normal force excluded the tension of the rope, I was going to find the difference in magnitude between the calculated normal force and the Y component of the 70 N (the tension) triangle to find the normal force including the tension of the rope.

    So I found the normal force (excluding the tension of the rope):

    FN = mg cos(θ)

    FN = 8.0 * -9.81 cos(40)

    FN = -78.48 cos(40)

    FN = -60.12

    Since the tension vector was the X component of the triangle I drew, I found the Y component of the triangle using trigonometry:

    tan(θ) = Opposite/Adjacent

    tan(50) = Opposite/70

    Opposite = tan(50) 70

    Opposite = 83.42 N

    Now that I had the Y component of the triangle, I found the difference between that and the calculated normal force:

    Normal Force Including Tension Vector = Calculated Normal Force + Y Component Of Tension Vector

    Normal Force Including Tension Vector = -60.12 + 83.42

    Normal Force Including Tension Vector = 23.3 N

    So I found that the normal force was 23.3 N, but when I checked my textbook the answer was 33.4 N :cry:

    Maybe I was overcomplicating things and horribly confused. I'm struggling with this question and it would be great if somebody could help me and tell me where I went wrong.

    I will post a picture of my diagram later.

    Also, this is my first post on PF and so I'm really excited! (And upset about the question).

    Thank you.
  2. jcsd
  3. Jan 24, 2013 #2


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    Are you sure there's a hill or inclined surface? From the question, it sounds like the ground is flat, and the rope makes an angle 40 with the horizontal.
  4. Jan 24, 2013 #3


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    Hi FredericChopin, welcome to Physics Forums!

    The statement of the problem does not mention a hill. To me it sounds like the sled is pulled along the horizontal ground with a rope that is tilted upward by 40o from the horizontal.

    [Edit: Seems I'm dx's echo tonight :smile:]
  5. Jan 24, 2013 #4

    I feel REALLY REALLY REALLY stupid. All that work for nothing... :frown:

    Thank you for clearing that up. But would one of you (or two of you) be able to do me a favour?

    Given that there was no hill, the ground was flat and the rope was making an angle of 40° to the horizontal, show, step by step, how you would solve this question.

    Thank you.
  6. Jan 25, 2013 #5


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    We are not allowed to provide step by step solutions in the homework forums, according to PF rules.

    But the problem is very easy. There are three forces acting on the block. The force from the rope, the weight of the sled, and the normal force from the ground.

    The normal force must balance the weight of the sled and the vertical component of the rope force, since the sled is not moving in the vertical direction. So start by finding the vertical component of the force exerted by the rope.
  7. Jan 25, 2013 #6


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    We aren't suppose to give step by step solutions. But if you will show your attempt, we will be glad to provide help. You'll need to start again with a good free-body diagram of the forces and then consider the components of the forces (especially the vertical components).

    However, I'm off to bed for now. Will check back tomorrow.
  8. Jan 25, 2013 #7
    I tried the question again and got the correct answer. Thank you. :smile:
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