Force, Find the Angle Given 2 Forces of Equal Magnitude, but Different Direction

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Homework Help Overview

The problem involves two forces of equal magnitude exerted by horses pulling on ropes attached to a tree stump. The resultant force is given as 1.51 times the magnitude of the individual forces, and the goal is to determine the angle between the two ropes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods of vector addition and the implications of dividing the forces into triangles. Some explore the use of trigonometric functions to relate the forces and resultant angle, while others express confusion over the correct approach and the use of specific functions like arcsin versus arccos.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the validity of their approaches. There is a recognition of differing interpretations of the problem, particularly regarding the angle calculation, and some participants express uncertainty about the professor's provided answer.

Contextual Notes

Participants mention the need to assign arbitrary values to the forces for calculation purposes and express frustration over conflicting results and methods. There is a noted emphasis on understanding the geometric relationships involved in the problem.

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Homework Statement


Two horses pull horizontally on ropes attached to a tree stump. Each horse pulls with a force of magnitude F. If the resultant force (vector R) has the magnitude R = 1.51 F, what is the angle (in degrees) between the two ropes?


Homework Equations





The Attempt at a Solution


Ry= 0+cos
Rx= 1+ sin

R= 1.51

sin=cos/1.54
tan= 1/1.54
angle= 33 degrees

33 * 2 = total angle for both of 66 degrees
 
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I'm on my 5th hour for this problem and still can't get it. I realize it's necessary to assign an arbitrary value of say 1 to F. That gives F=1 and R=1.51, but everything I've tried to do with that information has failed.
 
Using vector addition... We don't know whether the three vectors make a right triangle (when putting the two F's nose to tail), so it's easier to divide them into 2 triangles. We'll call the new segment G.

So now F becomes the Hypotenuse, G is Opposite, and 1.51/2 F is the adjacent (divided in half because it's now the adjacent on 2 triangles instead of just 1).

F sin(angle) = 1.51/2 F => divide out the F => arcsin(1.51/2) = angle = 49.025 degrees

That's just half of your angle though since you split it into two triangles
49.025*2 = 98.05 degrees
 
Last edited:
Allenman said:
Using vector addition... We don't know whether the three vectors make a right triangle (when putting the two F's nose to tail), so it's easier to divide them into 2 triangles. We'll call the new segment G.

So now F becomes the Hypotenuse, G is Opposite, and 1.51/2 F is the adjacent (divided in half because it's now the adjacent on 2 triangles instead of just 1).

F sin(angle) = 1.51/2 F => divide out the F => arcsin(1.51/2) = angle = 49.025 degrees

That's just half of your angle though since you split it into two triangles
49.025*2 = 98.05 degrees

The professor gave an answer of 81.9 degrees.
 
Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.
 
Allenman said:
Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.

I really appreciate the response, I initially did what you did and was told to rethink it and that it was 81.9. That's what's been making me go insane. I don't know why this method is wrong.
 
Allenman said:
Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.

You had it right and explained it right, but used arcsin instead of arccos. I get how to do this now, thank you so much for helping me finally understand this.
 

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