A uniform drawbridge must be held at a 37 angle above the horizontal to allow sh
ID: 2061985 • Letter: A
Question
A uniform drawbridge must be held at a 37 angle above the horizontal to allow ships to pass underneath. The drawbridge weighs 45,000 and is 14.0 long. A cable is connected 3.5 from the hinge where the bridge pivots (measured along the bridge) and pulls horizontally on the bridge to hold it in place.Part A
What is the tension in the cable?
Part B
Find the magnitude of the force the hinge exerts on the bridge.
Part C
Find the direction of the force the hinge exerts on the bridge.
Part D
If the cable suddenly breaks, what is the magnitude of the angular acceleration of the drawbridge just after the cable breaks?
Part E
What is the angular speed of the drawbridge as it becomes horizontal?
Please explain.
Explanation / Answer
a) The forces on the bridge are
Weight of bridge W, -y direction at 7.0 m from hinge (center of bridge)
Tension in the cable T, horizontal +x direction at 3.50 m from hinge
Vertical force from hinge Fy, +y direction
Horizontal force from hinge Fx, -x direction
sum of forces in X direction = 0
T - Fx = 0 ----- (1)
sum of forces in Y direction = 0
Fy - W = 0
Fy = W = 45000 N
moment about hinge = 0
-W*7*cos37 +T*3.5*sin37 = 0
45000*7*cos37 = T*3.5*sin37
T = 1.194x10^5 N
b) Fy = 45000 N , Fx = T = 1.19x10^5 N
F = sqrt(Fx^2 + Fy^2)
F = sqrt [ 45000^2 + 1.19x10^5^2 ]
F = 1.276x10^5 N
c) direction ( theta) = arctan(Fy/Fx)
theta = arctan(45000/1.19x10^5)
theta = 20.7 degrees above horizontal
d) torque = I*alpha
I = moment of inertia = M*L^2/3
M = mass of bridge
L = length
toqrue = W*7*cos37
W*cos37 = M*L^2*alpha/3 (W = Mg)
Mg*7*cos37 = M*L^2*alpha/3
alpha = 3*g*7*cos37/14^2
alpha = 3*9.8*7*cos37/14^2 = 0.839 rad/s^2
e) 37 degree = 0.7 rad
wf^2 = wi^2 + 2*alpha*s
wf^2 = 0+2*0.839*0.7
wf = 1.1 rad/s
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