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Question about 3rd law and normal force

  1. Sep 16, 2008 #1
    Hi,guys ..........i have a question in my mind always confused me just like newton 3rd law but this one is different.

    here is my question.
    I always confuse with the normal force and weight force which is mass * gravity.

    for example lets say a coin tossed straight upward,and reaches its highest point and falls back down again.

    1.here is my question when the coin is moving upward after it released. is it the force down or up?.................because i am confuse with the normal force and the weight force thats pulls the coin down.I know the normal force always be up and weight force be down.but how do we know the force is up and down. and also how do we know if it is decreasing and increasing.

    2. my second question how about if the coin at highest point . is the force will be down or up ..but my guess is 0 because the acceleration will be zero.



    thanks.
     
  2. jcsd
  3. Sep 16, 2008 #2
    Re: question

    "here is my question when the coin is moving upward after it released. is it the force down or up?.................because i am confuse with the normal force and the weight force thats pulls the coin down.I know the normal force always be up and weight force be down."

    First, the 'normal force' is only present on an object when the object is supported (as in resting on the floor) and is the force which prevents the force of gravity from accelerating the object down. It is also referred to as the 'support force'.

    So the coin tossed in the air has no normal force acting on it, only the force of gravity, which always acts downward.

    "my second question how about if the coin at highest point . is the force will be down or up ..but my guess is 0 because the acceleration will be zero."

    Love this question. Think of it this way: acceleration the rate of change of velocity. If the rate of change of velocity (acceleration) is zero, then when the coin reaches the highest point (where its velocity is instantaneously zero), its velocity stops changing, which means it would simply stay there and not start moving downward. The acceleration cannot be 0 at the highest point, even if the velocity is.
     
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