Solving Two-Horse Stump Pulling Problem: Find F2

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Homework Help Overview

The problem involves two horses pulling on ropes attached to a stump, with one known force (F1 = 1300 N) and an unknown force (F2). The net force is described as having a magnitude equal to F1 and making a 90-degree angle with it. Participants are exploring the implications of the forces' directions and the relationships between them.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the interpretation of the forces being horizontal and the implications of the angle of F2. There is a focus on the relationships between the forces in the x-y plane and how to approach solving for F2 using trigonometric relationships.

Discussion Status

Some participants have provided clarifications regarding the definitions of directions and the nature of the forces involved. There is ongoing exploration of how to correctly apply the equations of motion and force balance without assuming acceleration, indicating a productive dialogue on the problem setup.

Contextual Notes

Participants are questioning the definitions of angles and directions related to the forces, as well as the appropriateness of using certain equations in the context of this problem. There is mention of potential confusion due to the wording of the problem statement.

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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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