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Math Help - Static and Kinetic Friction-NEED HELP!!

  1. #1
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    Static and Kinetic Friction-NEED HELP!!

    1.) What would be the acceleration of a 19 kg aluminum block if it were being pulled with a horizontal applied force of 300 N?

    2.) In moving a 35..0kg desk from one side of a classroom to the other, a teacher finds that a horizontal force of 275 N is necessary to set the desk in motion and force of 195N is necessary to keep it in motion with constant velocity. What are the coefficients of (a) static friction and (b) kinetic friction between the desk and the floor?

    3.) A 40kg crate is at rest on a level surface. If the coefficient of static friction between the crate and the floor is 0.69, what horizontal force is required to move the crate.

    These problems I know how to do them but I get stuck...
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  2. #2
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    Quote Originally Posted by victorfk06 View Post
    ...

    2.) In moving a 35..0kg desk from one side of a classroom to the other, a teacher finds that a horizontal force of 275 N is necessary to set the desk in motion and force of 195N is necessary to keep it in motion with constant velocity. What are the coefficients of (a) static friction and (b) kinetic friction between the desk and the floor?

    3.) A 40kg crate is at rest on a level surface. If the coefficient of static friction between the crate and the floor is 0.69, what horizontal force is required to move the crate.

    These problems I know how to do them but I get stuck...
    Let  F_f denote the frictional force and N the force normal to the surface. Then

    F_f = \lambda \cdot N

    where \lambda denotes the friction coefficient.

    F_f = \lambda \cdot N~\implies~\lambda = \dfrac{F_f}N

    to #2:

    Plug in the values you know:

    \lambda_{stat} = \dfrac{275\  N}{35 \cdot 9.81\  N} \approx 0.8

    \lambda_{kin} = \dfrac{195\  N}{35 \cdot 9.81\  N} \approx 0.568

    to #3:

    The force normal to the surface is

    N = 40\ kg \cdot 9.81\ \frac{m}{s^2} = 392.4\ N

    Therefore the friction is:

    F_f = 0.69 \cdot 392.4\ N = 270.756\ N
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  3. #3
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    Thanks now I understand but still working and stuck on the first one, it seems very easy but dont know what Im not getting....
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  4. #4
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    Quote Originally Posted by victorfk06 View Post
    1.) What would be the acceleration of a 19 kg aluminum block if it were being pulled with a horizontal applied force of 300 N?
    a = \frac{F_{net}}{m}

    were you given any info regarding other forces besides the applied force (like friction, etc.) ?
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