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

In summary, the problem involves two horses pulling on ropes attached to a tree stump with a force of magnitude F. The resultant force (vector R) has a magnitude of 1.51 F. To find the angle between the two ropes, the vectors are divided into two triangles and using vector addition, the angle is calculated to be 49.025 degrees. However, the professor gave an answer of 81.9 degrees, which may have been calculated using arccos instead of arcsin.
  • #1
Simon777
35
0

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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  • #2
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.
 
  • #3
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:
  • #4
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.
 
  • #5
Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.
 
  • #6
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.
 
  • #7
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.
 

What is a force?

A force is a push or pull that can change the motion or shape of an object.

How do you calculate the angle between two forces of equal magnitude but different direction?

To calculate the angle between two forces, you can use the law of cosines. First, find the dot product of the two forces. Then, divide the dot product by the product of the magnitudes of the two forces. Finally, take the inverse cosine of the result to find the angle.

Can the angle between two forces be negative?

No, the angle between two forces cannot be negative. It is always measured as a positive value between 0 and 180 degrees.

How do you know which direction the angle is measured in?

The angle is measured in the counterclockwise direction from the first force to the second force.

What is the unit for force?

The SI unit for force is Newton (N). However, force can also be measured in other units such as pound-force (lbf) or dyne (dyn).

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