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Dynamics Problem help !

  1. Apr 8, 2005 #1
    Dynamics Problem...help plz!

    hey,
    i have a dynamics question here that i'm not quite sure how to solve:

    A stone weighs 5.4N. What force must be applied to make it accelerate at 3.0/s^2. (The answer is 7.1N must be applied upwards)

    ok, so here's what i've got so far:
    -gravity (9.81m/s^2) obviously takes part in this problem.
    -i'm thinking that the mass (in kg) must be found. i tried the formula Fg=mg to try and find this, however, i got a huge decimal number giving me doubt.

    help would be greatly appreciated!

    -jen
     
  2. jcsd
  3. Apr 8, 2005 #2
    theres many ways you can do this i'll show the second force law way

    taking the upward direction to be positive
    i this stone was moving upward with acceleration of 3.0m/s^2 then what is the force is required to make this stone accelerate at 3.0m/s^2??

    Fnet = ma (agree?)
    now, if a force of ma is making this stone move upward, then there must a force that is pushing this stone upward, which is in the opposite to the gravitational force taht is exerted by earth on the stone

    the sum of those two force would be Fnet = Fapp - mg
    but Fnet = ma soo
    ma = Fapp - mg
    Fapp = ma + mg
    mass,m should be in kilograms
     
  4. Apr 8, 2005 #3
    awesome! now i understand it :D
    ty very much.

    alright, if there's anyone who looks back at this post...i have one other physics problem that i just came across. the only reason why i'm having trouble with it is that it only gives me two pieces of information: the mass of the object and the coefficient of friction, mew. (i'll just use a "u" to show this"

    The driver of a 2200 kg car applies the brakes on a dry concrete road. Determine the force of friction between EACH tire and the concrete surface. u=0.60. (the answer given is 3.2kN)

    so i know that the formula for friction is: Ff=uFn
    i've also come up with: Fnet=ma
    =Ff-Fa
    =uFn-Fapp
    =umg-Fapp
    what i'm stuck on is how to calculate applied force with only the mass and mew. or am i missing something here? is there some other way to solve this problem?
     
    Last edited: Apr 8, 2005
  5. Apr 9, 2005 #4
    well ok sure there is a velocity but one thing WHO CARES!

    the force of friction (kinetic) stays the same regardless of the speed, right?

    Now since there are FOUR tires (tyres) each tire(tyre) must share the load of the car uniformly. So the load of the each tire (tyre) on the car must be mg/4, which is 1/4 the weight of the car. Find the normal force on each tire, and then find the frictional force.

    Kinetic Frictional force is [itex] F = \mu F_{N} [/itex], it doesnt depend on other forces since kinetic frictional force is constant over any given velocity
     
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