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Newton's Laws homework questions

  1. Sep 29, 2003 #1
    If anyone could help me with these problems, I'd be eternally grateful ...

    1) The distance between two telephone poles is 51.0 m. When a 0.800 kg bird lands on the telephone wire midway between the poles, the wire sags 0.210 m. How much tension does the bird produce in the wire? Ignore the weight of the wire.

    2) A block slides down a friction-less incline having an angle of 17.0 degrees. Assume that the block starts from rest at the top and the length of the incline is 2.60 m. Take the direction of motion as the positive direction. What's the acceleration of the block?

    3) A 21.0 kg block is initially at rest on a horizontal surface. A horizontal force of 78.0 N is required to set the block in motion. After it is in motion, a horizontal force of 59.0 N is required to keep the block moving with constant speed. Find the coefficients of static and kinetic friction from this information.

    4) A woman at an airport is towing her 20.0 kg suitcase at constant speed by pulling on a strap at an angle of (theta) above the horizontal. She pulls on the strap with a 35.0 N force, and the friction force on the suitcase is 20.0 N. What angle does the strap make with the horizontal? What normal force does the ground exert on the suitcase?
  2. jcsd
  3. Sep 29, 2003 #2


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    (1) Draw a picture. If you draw in the original position of the telephone wire and the new position, you will see a right triangle with horizontal leg of length 51/2= 25.5 m and vertical leg of length 0.21 m. You can use the Pythagorean theorme to calculate the length of the hypotenuse (the stretched length of the wire). The force diagram will be a "similar triangle" and since you know that the vertical force is mg (0.8*9.8) you can calculate the force in the direction of the wire which is its tension.

    (2) Another picture would help! You know the length (hypotenuse) of the incline and you know the angle so you can calclate the lengths of the horizontal and vertal legs. You know the downward component of acceleration is 9.8. Once again use "similar triangles" to find the acceleration in the direction of the incline.

    (3) Do you know the definitions of the coefficients of static and kinetic friction? The static friction is DEFINED as the force necessary to start the block moving and you are given that. The coefficient of kinetic friction is the fraction of the component of force perpendicular to the plane that is equal to the force necessary to keep the block moving.

    (4) Since the friction force is 20N she must be applying that amount horizontally to overcome the friction. She is applying 35 N force and that is the "hypotenuse" of the triangle. Use that to determine the angle. Of course, the vertical component (vertical leg of the triangle) is the Normal force.
  4. Sep 29, 2003 #3
    1) I understand everything you're saying about the triangles up to the point of finding the hypotenuse. So I found the hypotenuse to be 25.5 m. However, I'm a little confused on the similar triangle idea. Should I draw another triangle for force and how does it relate to the triangle I just drew? I just don't understand the last sentence.

    2) Alright, so I found the vertical leg to be .760 and the horizontal one to be 2.48. What does acceleration and similar triangles have to do with this? I guess I need a walk through on similar triangles.

    3) Static friction may be defined as that, but when I put 78.0 in the answer box, it's wrong. Also, I'm lost in the definition of kinetic force and how I use the answer from static force to obtain it.

    4) I got the angle right, but wouldn't the normal force of the triangle I drew be the normal force on the lady and not on the suitcase? Because the vertical component I found isn't working.

  5. Sep 30, 2003 #4


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    1 The point of drawing the triangles is that the forces point in those directions and are proportional to the lengths of the sides.
    You know the downward force is the weight of the bird. Divide that by the distance the wire sags to find the proportion. Multiply that proportion by the length of the hypotenuse to find the force along the wire- that's the tension in the wire.

    2 You calculate the accleration from "f= ma" once you know the force along the incline. You know the downward component of force: the weight of the block. You know the angle the incline (and therefore the force along the incline) make with the horizontal is 17 degrees.
    You know that "opposite/hypotenuse"= sin(17) = weight/(force along incline).

    3 I said "The static friction is DEFINED as the force necessary to start the block moving and you are given that." The COEFFICIENT of static friction is that force divided by the weight of the block.

    4 Error on my part. The normal you calculated is, as you said, the force on her (as well as the force she is exerting on the suitcase). The force of the ground on the suitcase is the weight of the suitcase minus that.
  6. Oct 8, 2003 #5
    I may be able to help for question #3

    In easier terms the kinetic friction of an object is its friction when the object is in motion. Static friction is the force of friction when an object just begins to move...the point where the applied force and friction force are equal.

    to determine the coeficient of ì

    ì is equal to the Friction Force divided by the mass times gravity


    = 78/21x9.8 (g=9.8)
    = 0.379008746
    Therefore this is the coefficient of Static Friction

    For the other you repeat the same thing except your friction force is 59 N rather than 78N....I hope that helps...Ill try to figure out the others but Im still learing my phyiscs also...lol
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