How Does Anchor Rode Angle Affect Tension and Force Mechanics?

  • Thread starter Thread starter Bad Monkey
  • Start date Start date
AI Thread Summary
The discussion centers on how the angle of an anchor rode affects the tension and force mechanics in a boat at anchor. When the rode is at a 45-degree angle, the tension consists of both horizontal resistance to wind force and a vertical component related to buoyancy. Participants explore whether this increased force can be classified as leverage or force multiplication, comparing it to scenarios involving weights suspended by ropes at angles. The concept of mechanical advantage is debated, with some suggesting it applies even in static situations, while others question its relevance. Overall, the conversation seeks to clarify the principles behind the observed force dynamics in these anchoring scenarios.
Bad Monkey
Messages
28
Reaction score
0
Just a basic question I am having trouble articulating.

A boat is at anchor with a single anchor and a rode with neutral buoyancy. The rode is of such a length and the anchor at such a depth that the rode is at 45 degrees to horizontal. The wind and current pushes horizontally on the boat with force Fw.

The anchor resists the pull such that it does not drag.

Because the rode is at 45 degrees, the tension in it must clearly be made up of the horizontal resistance to Fw plus the vertical component - Fw / cos(45).

My question is, how do we articulate this "increased" force in the rode. Is it a form of leverage, or force multiplication? If so what kind? Else what?
 
Physics news on Phys.org
Hi Bad Monkey! :smile:
Bad Monkey said:
My question is, how do we articulate this "increased" c. Is it a form of leverage, or force multiplication? If so what kind? Else what?

What increased force in the rope? :confused:

The horizontal component matches the wind, and the vertical component matches the increased buoyancy force caused by the boat being slightly lower.
 
Okay I know that, but the only external force acting on the system is the horizontal one from the wind. We end up with a summed force vector in the rode greater than that.

I suppose it is just like hanging a weight with two ropes at angles away from vertical - the summed forces of the two ropes is greater than the weight force. How do we articulate, in general, that force multiplication?
 
Hi Bad Monkey! :smile:

(why do you keep saying "rode"? :confused: it's a rope … if you're thinking of "rode at anchor", that's just the past tense of "ride at anchor" :wink:)

It's the same if it was a train on an embankment, anchored to the ground below by a rope at an angle θ …

if a wind, or the engine, pulls on the train with force F, the tension in the rope will be Fcosθ.
 
Bad Monkey said:
I suppose it is just like hanging a weight with two ropes at angles away from vertical - the summed forces of the two ropes is greater than the weight force. How do we articulate, in general, that force multiplication?

Making the example more specific:
A horizontal cable is strung between two anchor points, and then a weight is attached in the middle. The cable will sag a little, but not much, and the resulting tension in the cable is much larger than the force exerted by the weight.

I don't think there is a specific name for that, but obviously the way it works is the same as for a lever. In the case of a lever the ratio of the arm lengths determines the force ratio, and in the case of the cable there is a similar length-ratio and force-ratio relation. So 'lever action' or something like that will do.
 
Is it "mechanical advantage"?

I had thought not because I thought that applied only to machines, and I don't see how work is done in these static examples.
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top