Results 1 to 2 of 2

Math Help - friction and sliding box

  1. #1
    Senior Member
    Joined
    Jan 2007
    Posts
    476

    friction and sliding box

    I know there was a similiar problem i did awhile back involving true weight, that was elevator related, but it didn't include friction.

    also, i thought when an object is stationary then the Force was 0.

    friction is causing me some confusion here.

    A 6.40 kg box is sliding across the horizontal floor of an elevator.
    The coefficient of kinetic friction between the box and the floor is 0.460.

    SO, determine the kinetic frictional force that acts on the box for each of the following cases.

    (a) The elevator is stationary.
    in Newtons

    (b) The elevator is accelerating upward with an acceleration whose magnitude is 1.90 m/s^2.
    in Newtons

    (c) The elevator is accelerating downward with an acceleration whose magnitude is 1.90 m/s^2.
    in Newtons
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by rcmango View Post
    I know there was a similiar problem i did awhile back involving true weight, that was elevator related, but it didn't include friction.

    also, i thought when an object is stationary then the Force was 0.

    friction is causing me some confusion here.

    A 6.40 kg box is sliding across the horizontal floor of an elevator.
    The coefficient of kinetic friction between the box and the floor is 0.460.

    SO, determine the kinetic frictional force that acts on the box for each of the following cases.

    (a) The elevator is stationary.
    in Newtons

    (b) The elevator is accelerating upward with an acceleration whose magnitude is 1.90 m/s^2.
    in Newtons

    (c) The elevator is accelerating downward with an acceleration whose magnitude is 1.90 m/s^2.
    in Newtons
    You do all these in the same way as for a stationary lift, that is the frictional
    force is F=\mu_k m g, but here g is the effective accelaeration due to gravity,
    in the three cases this is g_0, g_0+1.9 and g_0-1.9, where g_0 is the conventional
    acceleration due to gravity g_0\approx 9.81 \mbox{m/s}^2

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sliding jigsaw configurations
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: July 26th 2011, 08:28 PM
  2. Sliding ladder problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 31st 2010, 07:22 PM
  3. Sliding Ladder (Related Rates)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 5th 2008, 11:09 AM
  4. physics-static friction + kinetic friction between blocks
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: February 20th 2007, 05:14 AM
  5. Friction and coefficient of friction questions!
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: March 17th 2006, 08:45 PM

Search Tags


/mathhelpforum @mathhelpforum