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    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
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    #1

    Oct 19, 2010, 08:45 AM
    Weight of block on incline vs frictional force
    Ok I know this is simple but I got 3 points taken off my last exam because it said:

    There is a block at rest on an incline, the frictional force acting on the block is...

    A)zero B)equal to the weight of the block

    C)greater than the weight of the block D)less than the weight of the block

    I said (C) greater than the weight of the block because if the friction on the block is greater than the weight wouldn't there not be enough force to make it move? The answer is (D). Why? Did I just get confused about how friction works?
    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
    Junior Member
    		  
    #2

    Oct 19, 2010, 05:56 PM

    I got it.

    Since the box is not slipping we can ignore fs(max)=un.
    or fs(max)=umgcos(a)

    Thus the fs=mgsin(a)

    Weight of box=mg

    mg>mgsin(a)

    Thanks kpg0001! Too late unknown, maybe next time.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
    Uber Member
    		  
    #3

    Oct 20, 2010, 09:37 AM

    Heh, there were some similar questions earlier on this month itself ;)
    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
    Junior Member
    		  
    #4

    Oct 20, 2010, 10:57 AM

    Yeah I was mistook the "weight of the block" for the normal force.
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