Solve Drawbridge Problem: Will Cable Break Before Lancelot Reaches End?

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The discussion revolves around a physics problem involving Sir Lancelot riding onto a drawbridge supported by a cable that will break under a tension of 5000 N. The total weight on the bridge, including Lancelot, his horse, and armor, is 5800 N, while the bridge itself weighs 1900 N. Participants explore the application of Newton's second law and torque to determine if the cable will break before Lancelot reaches the end of the bridge. Key calculations involve balancing torques around the hinge point of the bridge and considering the distribution of weight. The goal is to find out how far from the castle Lancelot will be when the cable breaks, emphasizing the importance of understanding forces and distances in torque calculations.
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Homework Statement



1.Sir Lancelot rides out of the castle at Camelot and onto the 12.0 m long drawbridge that passes over the moat. Unbeknownst to him, his enemies have partially severed the cable (60° angle with the horizontal) holding up the front end of the bridge so that it will break under a tension of 5000N. The bridge weighs 1900 N and it is uniform. Lancelot, his lance and armor and his horse have a combined weight of 5800 N. Will the cable break before Lancelot reaches the end of the drawbridge? If so, how far from the castle will he be when the cable breaks? (measure from the castle to the center of mass)


The Attempt at a Solution




looking at this I am not really too sure. I may have to use Newtons 2nd law to set up equation. Not sure how to possibly go about finding the distance away he would be when the cable broke.
 
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It's a torque question.
Starting from the hinge work out the force * distance for each component.
Weight of bridge * middle + weight knight * distance downward = force on ropes vertically * length.
 
ok...I don't really understand where the distances come from or how to apply any formula to this problem.
 
Torque is force (at right angles) * distance - eg. when something is trying to turn.
Then it just like balancing forces on a see-saw (teeter-totter)
So you just pick a point to calculate the torques about - it doesn't matterwhere but the axle is an obvious place.

For the weight of the bridge you could imagine summing the weight of each tiny area at a distance X from the hinge - or it's fairly obvious that this equals the weight * half the length, ie the weight acting at the mid point.

Then you also have the weight of the horse + knight at a distance 'd'

To balance these two downward torques you must have an upward one - the ropes are acting at the end of the bridge so the distance is the full length of the bridge.
 
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