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Find the angular acceleration of the bridge?

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed in his quest to improve communication between damsels and dragons (Fig. P12.20). Unfortunately his squire lowered the draw bridge too far and finally stopped it 20.0° below the horizontal. Lost-a-Lot and his horse stop when their combined center of mass is 1.00 m from the end of the bridge. The uniform bridge is 8.00 m long and has a mass of 2100 kg. The lift cable is attached to the bridge 5.00 m from the hinge at the castle end, and to a point on the castle wall 12.0 m above the bridge. Lost-a-Lot's mass combined with his armor and steed is 1000 kg. Suddenly, the lift cable breaks! The hinge between the castle wall and the bridge is frictionless, and the bridge swings freely until it is vertical.

    Find the angular acceleration of the bridge once it starts to move.
    theta = 20deg
    mb = 2100 kg
    mt = 1000 kg
    d1 = 1m (length from knight to end of drawbridge)

    2. Relevant equations
    torque τ=RxF= Iα

    3. The attempt at a solution

    I don't think the hinge force is relevant here and the tension force certainly isn't since it has snapped.

    Therefore, the only forces causing torque are those of the mass-gravity force of the bridge and the night.

    I calculate these as as Fb= mb*gsin(θ) and Ft = mt*gsin(θ) respectively; these forces are applied perpendicular to the drawbridge at 4 meters and 7 meters from the hinge respectively.

    Therefore, the total torque as given by RxF should be 4Fb + 7Ft.

    I equate this to Iα in order to solve for the angular acceleration denoted by alpha, meaning that
    α=τ/I = RxF/I

    The moment of inertia, I, was calculated by I = 1/3*mb*L2 + mt(L-d1)^2.

    My answer once everything is plugged in:

    (4*2100*9.81sin(20deg) + 7*1000*9.81sin(20deg)) / (1/3*2100*8^2+1000*7^2) = 0.5509 rad/s^2 physics.png
  2. jcsd
  3. Apr 18, 2017 #2


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    Incorrect. The forces are straight down because they are the force of gravity on the drawbridge and on the knight. You need to calculate the lever arm for the torques.
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