Results 1 to 5 of 5

Math Help - Mechanics question

  1. #1
    Banned
    Joined
    Mar 2011
    Posts
    118

    Mechanics question

    A dog lies on the floor and refuses to move. When you pull the lead with a force
    of 100 N and at an angle of 30 with the horizontal, the dog slides along the floor.
    Determine the coefficient of friction, if the dog has a mass of 25 kg. (You can take

    g = 10 m/s2.)

    My working. [LaTeX ERROR: Convert failed]
    Surely I need to have been told the acceleraion to work out F.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by poirot View Post
    A dog lies on the floor and refuses to move. When you pull the lead with a force
    of 100 N and at an angle of 30with the horizontal, the dog slides along the floor.
    Determine the coefficient of friction, if the dog has a mass of 25 kg. (You can take

    g = 10 m/s2.)

    My working. [LaTeX ERROR: Convert failed]
    Surely I need to have been told the acceleraion to work out F.
    \mu_s = \dfrac{f_{s \, max}}{F_N}

    \mu_s = \dfrac{100\cos(30^\circ)}{25g - 100\sin(30^\circ)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,848
    Thanks
    321
    Awards
    1
    Quote Originally Posted by poirot View Post
    My working. [LaTeX ERROR: Convert failed]
    skeeter is correct, but I am concerned about this line above.

    Forces are vectors so when you calculate the net force you are doing a vector sum. The simplest way to do this is to add the forces by components. In your work it looks like you just simply tried to add all the forces. This is not a correct approach.

    I will use the usual xy coordinate system and say that the dog is being pulled up and to the right. In the FBD we have a weight w acting straight down (in the -y direction), a normal force F_N acting directly upward, an applied force F acting up and to the right at a 30 degree angle, and a friction force f acting to the left (in the -x direction.)

    \sum F_x = F~cos(30) - f = ma

    Since we are looking for the coefficient of friction we are going to set a = 0 m/s^2. This gives us the maximum possible static friction, which is given by f = \mu F_N. So
    \sum F_x = F~cos(30) - f = 0

    In the y direction:
    \sum F_y = N + F~sin(30) - mg = 0

    Solve this equation for F_N and solve the x equation for f and you can then apply skeeter's solution.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Mar 2011
    Posts
    118
    Quote Originally Posted by topsquark View Post
    skeeter is correct, but I am concerned about this line above.

    Forces are vectors so when you calculate the net force you are doing a vector sum. The simplest way to do this is to add the forces by components. In your work it looks like you just simply tried to add all the forces. This is not a correct approach.

    I will use the usual xy coordinate system and say that the dog is being pulled up and to the right. In the FBD we have a weight w acting straight down (in the -y direction), a normal force F_N acting directly upward, an applied force F acting up and to the right at a 30 degree angle, and a friction force f acting to the left (in the -x direction.)

    \sum F_x = F~cos(30) - f = ma

    Since we are looking for the coefficient of friction we are going to set a = 0 m/s^2. This gives us the maximum possible static friction, which is given by f = \mu F_N. So
    \sum F_x = F~cos(30) - f = 0

    In the y direction:
    \sum F_y = N + F~sin(30) - mg = 0

    Solve this equation for F_N and solve the x equation for f and you can then apply skeeter's solution.

    -Dan
    Thanks a lot for your help
    I see my error, I was taking the friction to go up the slope so was resolving forces paralell to the slope. So frictional co-efficent = max friction/normal reaction? What does frictional co-efficent really mean ? I got sqrt(3)/4 by the way
    Last edited by poirot; March 13th 2011 at 03:14 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,848
    Thanks
    321
    Awards
    1
    Quote Originally Posted by poirot View Post
    Thanks a lot for your help
    I see my error, I was taking the friction to go up the slope so was resolving forces paralell to the slope. So frictional co-efficent = max friction/normal reaction? What does frictional co-efficent really mean ? I got sqrt(3)/4 by the way
    Yes on both counts. The coefficient of (static in this case) friction is kind of an average of rough surfaces, "stickiness" of intermolecular forces, etc.) It's kind of messy when you get down to what causes it. But for (so far as I know of) all pairs of surfaces we can take the maximum static friction force to be some coefficient \mu _s times the normal force. (Kinetic friction is always \mu _k~F_N.) The larger the coefficient of friction the harder it is to make the surfaces move across each other.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mechanics Question
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: March 8th 2010, 12:33 PM
  2. mechanics question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: March 1st 2010, 09:27 AM
  3. C.O.M Mechanics question.
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: June 14th 2009, 02:48 PM
  4. Another mechanics question.
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: June 13th 2009, 12:46 AM
  5. Mechanics question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 4th 2008, 09:47 AM

Search Tags


/mathhelpforum @mathhelpforum