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Magnitude of the force formula

  1. Feb 1, 2004 #1
    I've got a lot of home work to do still and my book has run out of formulas for the questions my Professor so kindly gave us.

    I'm not the brightest crayon in the box, and I know you arn't here to do my home work for me. so I'm just asking for some formuals, or somewhere to find formuals to find

    average force when I'm only given two speeds and the amount of contact time between a ball and bat.

    highest point-given two velocities, its on an angle (a skier jumping a ramp) but it doesn't give any information about the angle, or how long the ramp is etc.

    kenitic friction on ice for minimun initial speed- a hockey player hit a puck across come ice and it only went half way. Given the initial speed and that it only reaches half way. I'm supposed to figure out the initial speed he should have given the puck to get it to reach the other player.

    Finially I keep seeing "magnitude of the force" and I cant remember what this means or what its asking me and I'm getting really frusterated !!

    please try to help me. Thank you!
  2. jcsd
  3. Feb 1, 2004 #2
    Now force = Change in momentum per unit time

    Highest point will be achieve when the velocity in upward firection would be zero

    Force is a vector quantity so it has got magnitude u shoul also look for definition of vectors in this regards
  4. Feb 1, 2004 #3


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    For the first one, you should use the impulse-momentum theorem:
    [tex] \vec{I} = \Delta \vec{p} [/tex]
    The impulse can be found by multiplying a constant average force by the time interval.

    I'm not clear on the second question. Are you given a horizontal and vertical velocity? If so, that (along with the assumption of no air resistance) will uniquely define a trajectory. You should have kinematic equations for rigid body motion with constant acceleration that you can use to find the highest point in the trajectory.

    If you know the initial speed of the hockey puck and how far it went, you can find the friction: assume the only force acting on the puck after it is hit is the (constant) kinetic friction, f. Let the subscripts 1 and 2 represent the situation where the puck is hit halfway and all the way, respectively. Here's one way to solve the problem using the work-energy theorem:

    [tex] 0 - \frac{mv_{o1}^2}2 = -f d_1 [/tex]
    [tex] 0 - \frac{mv_{o2}^2}2 = -f d_2 [/tex]

    Solve the first equation for the friction force, f, and plug it into the second equation, which you can solve for vo2, the minimum initial velocity needed to reach the other hockey player. Note that d2 = 2*d1

    For your last question, this goes back to the definition of a vector. Force, like velocity and acceleration and many other physical quantities, is a vector quantity. Vector quantities are specified by both their "magnitude" and "direction." When you draw a graphical reperesentation of a vector, the direction is the angle between the vector and some specified reference axis and the magnitude is the length of the vector (multiplied by whatever scale factor you are using).

    Hope that helps.

    EDIT: Oops, I didn't check to see if this was replied to already.
    Last edited: Feb 1, 2004
  5. Feb 1, 2004 #4
    Heres the questions I haven't answered yet. For you're giggling pleasure. It might help you help me more. I'm still lost. I've only been working with all these terms and stuff for two weeks now so big words and symbols confuse me more

    A water-skier lets go of the tow rope upon leaving the end of a jump ramp at a speed of v1 = 16.1 m/s. As the figure below indicates, the skier has a speed of v2 = 13.1 m/s at the highest point of the jump.
    Ignoring air resistance, determine the skier's height H above the top of the ramp at the highest point.

    In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 1.96 m/s. However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice. As a result, the puck travels only one-half the distance between the players before sliding to a halt. What minimum initial speed should the puck have been given so that it reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?

    A baseball (m = 146 g) approaches a bat horizontally at a speed of 39.4 m/s (88.1 mph) and is hit straight back at a speed of 48.6 m/s (109 mph). If the ball is in contact with the bat for a time of 1.18 ms, what is the average force exerted on the ball by the bat? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

    The speed of a bobsled is increasing, because it has an acceleration of 2.32 m/s2. At a given instant in time, the forces resisting the motion, including kinetic friction and air resistance, total 447 N. The mass of the bobsled and its riders is 286 kg. What is the magnitude of the force propelling the bobsled forward?

    A falling skydiver has a mass of 118 kg. What is the magnitude of the skydiver's acceleration when the upward force of air resistance has a magnitude that is equal to one-fourth of his weight?

    A 72.8 kg water skier is being pulled by a horizontal force of 521 N and has an acceleration of 2.40 m/s2. Assuming that the total resistive force exerted on the skier by the water and the wind is constant, what force is needed to pull the skier at a constant velocity?

    The Deep Space Probe's mass is 476 kg. At a thrust of 54.8 mN how many days are required for the probe to attain a velocity of 785 m/s (1756 mph), assuming that the probe starts from rest and that the mass remains nearly constant?

    The figure below shows three particles far away from any other objects and located on a straight line.
    The masses of these particles are mA = 353 kg, mB = 533 kg, and mC = 160 kg. d1 = 0.478 m and d2 = 0.239 m. Calculate the magnitude of the net gravitational force acting on particle A.
    Last edited: Feb 1, 2004
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