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    Maria A Gonzale's Avatar
    Maria A Gonzale Posts: 13, Reputation: 1
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    #1

    Sep 29, 2008, 08:00 PM
    Right direction physics
    If 460 kg bear grasping a vertical tree slides down at constant velocity. With acceleration of gravity being 10 m/s^2. What is the fraction force that acts on the bear?
    Will I use F=ma, so 460 times 10?
    Am I in the right direction
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Sep 30, 2008, 05:39 AM

    Yes, you are on the right track. If the bear slides at constanjt velocity, then you know that its acceleration is 0. Hence F = 0. This means that the bear's weight is exactly counter-acted by the friction of the bear's claws against the tree. Can you take it from here?
    Maria A Gonzale's Avatar
    Maria A Gonzale Posts: 13, Reputation: 1
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    #3

    Sep 30, 2008, 02:06 PM

    So will the frictional force be 460N
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Sep 30, 2008, 02:16 PM

    Not quite. The bear's mass is 460 kg, but his weight is a force, given in Newtons. If you know an object's mass, its weight under earth gravity is mg.
    Maria A Gonzale's Avatar
    Maria A Gonzale Posts: 13, Reputation: 1
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    #5

    Oct 1, 2008, 01:32 PM

    Okay I am confused, so exactly what am I doing wrong...
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Oct 1, 2008, 03:55 PM

    Quote Originally Posted by Maria A Gonzale View Post
    okay i am confused, so exactly what am i doing wrong...
    Well... a 460Kg bear has a weight of 460kg * 10m/s^2 = 4600 Newtons.
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