How to Solve a Static Forces Angle Problem with Variable Length

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Homework Help Overview

The discussion revolves around a static forces problem involving a pulley system with variable lengths and angles. The objective is to determine the height (h) for different masses at point B, while considering the forces acting on the system and the relationships between angles and lengths.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationships between forces and angles, questioning how to effectively link equations involving unknown angles. There are attempts to express the static equilibrium conditions and explore the implications of the forces acting on the blocks.

Discussion Status

The discussion is ongoing, with participants exploring various approaches to relate the angles and forces. Some guidance has been offered regarding focusing on the components of forces, but there is no consensus on a specific method or solution yet.

Contextual Notes

Participants note the presence of multiple unknowns and express uncertainty about the necessary formulas and substitutions needed to progress. There is an acknowledgment of the complexity introduced by the angles and the static nature of the problem.

Madcat
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Homework Statement



I drew up a little diagram to illustrate the problem.

PhysicsLabQuestion.jpg


The goal of this lab report is to find the h value (in cm) for each of the possible masses at B. (one problem for 500g, one for 700g and one for 900g. The only thing that should change is the h value. The leftmost mass never changes)
θ and a are unknown. The pulley on the left is considered to be frictionless and the cords masses are negligible.

Point B is "tied" so the cord does not slide and thus BC is fixed at 60 cm. Length AB however is variable due to the pulley.

Homework Equations



Sin(θ) = opposite/hypotenuses
Cos(θ) = adjacent/hypotenuses
Tan(θ) = opposite/adjacent

The Attempt at a Solution



Here are the equations I've come up with thus far.

Cos(a) = ABx/4.9N
Sin(a) = ABy/4.9N
ABx/Cos(a) = ABy/ Sin(a)

cos(θ) = (105-d)/60 (d being the top side of the left-most triangle, separated at h)
tan(a) = h/d
h = sqrt( (60^2) - ((105-d)^2) )

I can also say that:
BCx = ABx (as this is a static question. B is at rest, thus all resulting forces acting on it are equal to 0)
BCy + ABy - W = 0
or alternatively
BCy + ABy = W
(where W is the current mass of the object at point B)

I "know" the solution lies in the angles, but for the life of me I just can't fit my equations together in order to draw a definite conclusion. I'd love a push in the right direction if anyone can contribute something to this. I keep trying to break down the forces but without an angle it is difficult and I can't find the proper substitution.

First time post here, I hope I did it right, thanks in advance for any help.
 
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Madcat said:
I "know" the solution lies in the angles, but for the life of me I just can't fit my equations together in order to draw a definite conclusion. I'd love a push in the right direction if anyone can contribute something to this. I keep trying to break down the forces but without an angle it is difficult and I can't find the proper substitution.

It seems that you're on the right track...

Take a step back before trying to solve for the angle. What are the forces acting on the two blocks? How do these forces relate to the angles?

-Kerry
 
Last edited:
It has been a week and I still seem to be unable to get the answer myself.
A colleague did the calculations and passed them to me, but they seem to essentially be my own calculations and don't give an h value (which we are searching for)
I plugged them into Maple 12 and it cannot solve it. I am quite certain I have too many unknowns and I cannot figure out how to elemenate them.

If someone could at least confirm I am not missing any formulas or maybe which angle to focus on? (I would figure I need to 'link' the formula's using the angles, but have yet to figure out how)

Thanks again.

Oh, and the formula's he gave me are...

-4.905*cos(a) - Tr cos(q)=0

-4.905 + 4.905*sin(a)+Tr*sin(q)=0

.6/(sin(a))= 1.05/(sin(180 - (a +q))
 
Last edited:
Welcome to PF!

Hi Madcat! Welcome to PF! :smile:

You know T, and you know that tan(a) = h/(105 - 60cosθ).

You don't know the tension in the 60 string, and you don't want to know it, so … standard trick … take components perpendicular to that string.

What do you get? :wink:
 

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