engineering science

An object weighing 60N rest on a slope forming an angle of 17.5 degrees with the horizontal. A force of 44N parallel to the sloping surface just moves the object upwards. Calculate the frictional force.

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asked by olivia
  1. break the 44 N force into components up the slope, and normal to the slope.
    F=44cos17.5 u + 44sin17.5 n

    now break the weight in to normal, and up the slope components.

    60N down= 60cos17.5 n - 60 sin17.5 u

    frictional force opposes upward, so it is a negative force up the slope.

    Now consider just the n forces.
    friction=mu(60cos17.5+44sin17.5)
    well, at this point, you are stuck without mu. So,
    compute that

    then up the slope...
    fu=friction
    44cos17.5 u-60 sin17.5=friction
    now you can compute friction, and go back with the normal equation, and compute mu.

  2. m*g = 60 N. = Wt. of object.

    Fp = 60*sin17.5 = 18.04 N. = Force
    parallel to the incline.

    Fn = 60*cos17.5 = 57.22 N. = Normal =
    force perpendicular to the incline.

    !@#$%^&-Fp-Fs = m*a
    44-18.04-Fs = m*0 = 0
    Fs = 44-18.04 = 26 N. = Force of static
    friction.

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    posted by Henry

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