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5 Quick Questions

  1. Nov 14, 2003 #1
    1) A father and his seven year old daughter are facing each other on ice skates. With their hands, they push off against one another. (a) COmpare the magnitudes of the pushing forces that they experience. b (b) Which one, if either, experiences the larger accceleration.

    *I am thinking that since F = M x A, the father will exert a greater force onto the child. And since the mass of the father is larger the child will have a greater acceleration.

    2) The force of air resistance acts to oppose the motion of an object moving through the air. A ball is thrown upward and eventualy returns to the ground. (a) As the ball moves upward, is the net force that acts on the ball greater than, less than, or equal to its weight? Justify your answer. (b) repeat part (a) for the downward motion.

    * I am thinking when the ball is moving up the mass of the ball is greater than the net force acting on it. And when it is moving down it is less? (Not sure please help)

    3) A stack of books whose true wight is 165N is placed on a scale in an elevator. the scale reads 165N. Can you tell from this information whether the elevator is moving with constant velocity of 2 m/s upward or 2m/s downward or whether the elevator is at rest? Explain.

    * It is at rest. Because if it is moving upward the books will eb heavier and if it moves downward they will be lighter. (again not sure please help)

    4) A rescue helicopter is lifting a man (weight- 822 N) from a capsized boat by the means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of 1.10m/s^2? (b) What is the tension during the remainder of the rescue when he is pulled upward at constant velocity?

    *I know that tension = mass * gravity but i Don't know how to do it with the acceleration.

    5) Three forces act on a moving object. One force has a magnitude of 80.0N and is directed due norht. ANother has a magnitude of 60.0N and is directed due west. What must be the magnitude and direction of the third force, such as the object continues to move with constant velocity.

    * I am thinking of using the pythagorean theorem.

    Any help is aprreciated. Please explain how to use each formula professor said this is going to be on the test and I am confused by it. Thanks!
  2. jcsd
  3. Nov 14, 2003 #2


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    The forces will be equal, by Newton's third, since they are opposite and no other forces come into the problem. So the girl experiences force F, say and the father experiences the opposite and equal force -F. Assuming the mass of the father is greater than the mass of the girl (this isn't a given - the father could be a midget and the daughter have accelerated growth), but assuming that then F = dad_mass X dad_accel = girl_mass X girl accel, Dividing by dad_mass we get dad_accel = (girl_mass/dad_mass) X girl_accel. Our assumption was that (girl_mass/dad_mass) was less than 1, so dad_accel is less than girl_accel.
  4. Nov 14, 2003 #3
    No: Newton's third law.

    Remember that F, M, and A in that equation refer to the same body. You can't stick the mass of the father and the acceleration of the child into the equation; if you want the acceleration of the child, you have to use the mass of the child and the force on the child.

    The mass of the ball? Do you mean the weight? (Weight is a force; mass isn't.)

    To answer this question, just note that the net force is the weight plus the frictional force; whether the net force is greater in magnitude than the weight, you have to determine whether the frictional force is adding to or subtracting from the weight (i.e., whether they point in the same or opposite directions).

    No, they will be heavier or lighter if they are accelerating upward or downward. But they are not accelerating, they are moving with constant velocity.

    No, tension isn't mass * gravity, (i.e., it's not weight). The tension is equal to the weight only if the man is not accelerating (i.e., in part (b)).

    You need to do it this way:

    Fnet = ma = T-w

    That will work for the magnitude, but not the direction. The direction you have to get by trigonometry (if they're looking for the angle).
  5. Nov 14, 2003 #4
    So for #2, you said it depends if the frictional force is adding or subtracting. If the ball is thrown up it is adding to the weight so it would be positive f. But when it is falling it is subtracting.

    For #3. You are not able to tell whether the object is moving at those velcoities because they are constant. Same goes for rest. If they were accelerating then it would be different?

    For #4. I know for part b the tension will be 822N but still don't know how to figure out the acceleration.

    For #5. Which trig function? Are you talking about cos and sin/

    Thanks again!!!
  6. Nov 14, 2003 #5
    Your statements concerning #2 and #3 are correct.

    If his velocity is constant, his acceleration is zero, by definition.

    If x is the x-component of the net vector and y is the y-component, and the vector has an angle θ (measured counterclockwise from the x axis), then tan(θ) = y/x.
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