A draw bridge of length L is being held up by an axel at its pivot and by a chai

Question

A draw bridge of length L is being held up by an axel at its pivot and by a chail that is also of length L. The drawbridge has uniform mass.

A) What is the tension in the chain as a function of the angle is makes with the horizontal?

B) Are there any angles that are mathematically impossible? Do these mathematical impossibilities make physical sense?

C) The drawbridge covers a moat of width L, protecting a castle wall that is H high. What is the minimum angle to shoot an arrow so that it only barely clears the wall?

Explanation / Answer

A) You can write the torque formula for the bridge as mgL/2 = Tsin(theta)*L. SOlving this for T, you should get mg/(2sin(theta)).
B) This would be mathematically impossible at 0 and 180 degrees - basically when the bridge is straight up or straight down which are possible in real life but would never really be done - we don't make bridge like that.
C) H = Vosin(theta)t-.5gt^2 and L = Vo cos(theta)*t. Now at maximum height, v = 0 = Vosin(theta)-gt so t = Vosin(theta)/g. Plugging this into the H equation, you should get H = (Vo^2 *(sin(theta))^2)/2g. Plugging into the L equation, you should get L = vo^2*cos(theta)sin(theta)/g. SOlving for Vo^2, we get Lg/(cos(theta)sin(theta)) = Vo^2. Plug this Vo^2 back into the H quation and simplify to get H = Lsin(theta)/(2cos(theta)). This then becomes 2H/L = tan(theta) and theta is just the arctan(2H/L)

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