Solving Two-Horse Stump Pulling Problem: Find F2

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The discussion centers on solving for the force F2 exerted by two horses pulling a stump, with F1 given as 1300 N and the resultant force R at a 90-degree angle to F1. The confusion arises from the interpretation of the forces being horizontal, clarified as the ropes being parallel to the ground. The solution involves using trigonometric relationships, specifically sine and cosine, to relate the forces in the x-y plane. Participants suggest focusing on force equilibrium rather than acceleration, emphasizing the importance of defining directions for clarity. Ultimately, the correct magnitude of F2 is determined to be 1838 N at an angle of 135 degrees.
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Homework Statement



Two horses pull horizontally on ropes attached to a stump. The two forces F1= 1300 N and F2 that they apply to the stump are such that the net (resultant) force R has a magnitude equal to that of F1 and makes an angle of 90 with F1.

What is the magnitude of F2?

Homework Equations


The Attempt at a Solution



I got the answer correct but I was confused with the wording. F2 ended up being at some angle with respect to the x-axis and F1. I thought that this force was horizontal. Could somebody clarify this? Also could this be solved using the dot product?

θ=135
F2= 1838N
 
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Toranc3 said:
I got the answer correct but I was confused with the wording. F2 ended up being at some angle with respect to the x-axis and F1. I thought that this force was horizontal. Could somebody clarify this?
When they say that the horses pulled horizontally, they just mean that the ropes were parallel to the ground, not angling upward or downward.

If F1 is along the y-axis, then the resultant will be along the x-axis. You use that fact to solve for F2. (The x-y plane is the plane of the ground.)

How did you solve it?
 
Doc Al said:
When they say that the horses pulled horizontally, they just mean that the ropes were parallel to the ground, not angling upward or downward.

If F1 is along the y-axis, then the resultant will be along the x-axis. You use that fact to solve for F2. (The x-y plane is the plane of the ground.)

How did you solve it?


Fy:

F=ma

ma=R
F2sin(theta) =1300N


Fx:

F=ma
F1+F2cos(theta) =ma

ma=0

-F1/cos(θ) =F2

I substituted this into the Fy. I am not sure if this is the correct way to do it though.
 
Toranc3 said:
Fy:

F=ma

ma=R
F2sin(theta) =1300N


Fx:

F=ma
F1+F2cos(theta) =ma

ma=0

-F1/cos(θ) =F2

I substituted this into the Fy. I am not sure if this is the correct way to do it though.
Please define your directions. Where is F1 pointing? What's θ?

Instead of F = ma, just use ƩF = whatever. There's no acceleration involved, just adding forces.
 
Doc Al said:
Please define your directions. Where is F1 pointing? What's θ?

Instead of F = ma, just use ƩF = whatever. There's no acceleration involved, just adding forces.

Yeah I did it over again with just forces. Thanks for your help.
 
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