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Thread: Friction question

  1. #1
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    Friction question

    The diagram shows a case of mass 15kg being dragged along a rough horizontal floor by means of a strap inclined at an angle A to the horizontal, where sinA = 0.6

    The tension in the strap is 50N. the coefficient of friction between the case and the floor is 0.25




    Modelling the case as a particle,

    a) show that the normal reaction between the case and the floor has magnitude 117N,

    b) find, to two decimal places, the acceleration of the case.



    I have tried F=0.25*(15*0.8) + 50*0.6

    this is Friction = coefficient*reaction + 50sinA
    Ive no idea if this is the correct thing to do though. I get friction=66.75.

    What do i do next though??
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  2. #2
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    equilibrium in the vertical direction ...

    normal force + vertical component of applied force = weight

    $\displaystyle F_N + 50(.6) = 15g$

    $\displaystyle F_N = 15g - 30 = 117 \, N$


    net force = horizontal component of the applied force - friction force

    $\displaystyle 15a = 50(.8) - 0.25(117)$

    $\displaystyle a = \frac{10.75}{15} = .72 \, m/s^2
    $
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  3. #3
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    Quote Originally Posted by djmccabie View Post
    The diagram shows a case of mass 15kg being dragged along a rough horizontal floor by means of a strap inclined at an angle A to the horizontal, where sinA = 0.6

    The tension in the strap is 50N. the coefficient of friction between the case and the floor is 0.25




    Modelling the case as a particle,

    a) show that the normal reaction between the case and the floor has magnitude 117N,

    b) find, to two decimal places, the acceleration of the case.



    I have tried F=0.25*(15*0.8) + 50*0.6

    this is Friction = coefficient*reaction + 50sinA
    Ive no idea if this is the correct thing to do though. I get friction=66.75.

    What do i do next though??
    Use a free body diagram and divide the forces up into horizontal and vertical components. I shall enumerate the respective components as follows:

    Vertical components:
    1. Normal force (upwards): $\displaystyle F_N = N$
    2. Gravity (downwards): $\displaystyle F_g = mg$
    3. Vertical component of the tension (upwards): $\displaystyle T_v = T sin \theta$

    Note that the vertical components must add up to zero net force, because the block is not accelerating upwards or downwards (not levitating nor sinking into the floor). In other words, the sum of the upward forces is equal in magnitude to the sum of the downward forces. Therefore, your first equation must say: $\displaystyle F_g = F_N + T_v \longrightarrow mg = N + T sin \theta$

    Horizontal components:
    1. Horizontal component of the tension (towards rope): $\displaystyle T_h = T cos \theta$
    2. Force of friction (against rope, as friction always opposes the direction of motion): $\displaystyle F_f = \mu N$

    Note that the horizontal components is what gives the block its acceleration - the block is moving in a horizontal direction towards the original of the tension/pull. Therefore, the net force on the block is expressed by the difference between the horizontal tension and friction: $\displaystyle F_{net} = T_h - F_f = T cos \theta - \mu N$

    Now, solve the two equations constructed:
    #1: $\displaystyle mg = N + T sin \theta$
    #2: $\displaystyle F_{net} = T cos \theta - \mu N$

    From the first equation, you will find that indeed $\displaystyle F_N = N = 117.15$ Newtons.

    To find $\displaystyle cos \theta$, use the identity $\displaystyle cos^2 \theta + sin^2 \theta = 1$, which will tell you that $\displaystyle cos \theta = 0.8$. Then your net force $\displaystyle F_{net} = T cos \theta - \mu N = 50 (0.8) - 0.25 (117.15) = 10.71$ Newtons. Using $\displaystyle F = ma$, we see that the acceleration is: $\displaystyle 0.714 \frac{m}{s^2}$

    In general, when solving physics problems, always try to use variables to represent your quantities. Try not to plug in numbers until the very end. This makes it easier to check over your own work if you make a mistake somewhere. It is easier to track what you're doing if you are, for example, swapping around the quantity $\displaystyle F_{fr} = \mu N = \mu mg$ than if you are just swapping around the number 3.14159 Newtons (then you don't know what it's supposed to represent).
    Last edited by Last_Singularity; Dec 29th 2008 at 10:28 AM.
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