Converting Force to Velocity: Tips for Solving Tricky Physics Problems"

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Converting force to velocity involves understanding the relationship between acceleration, time, and mass. The discussion centers on a physics problem where the final velocity of the center of mass is given, and the force for a third particle needs to be determined. Participants suggest using Newton's second law to break down forces into x and y components, allowing for the formulation of equations to solve for the unknown force. A user initially misapplied trigonometric functions but corrected their mistake to arrive at the correct answer. Accurate significant figures are also highlighted as a consideration in problem-solving.
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I am going brain dead! How do I convert force to velocity? I do know the acceleration and time and mass, but my conversion is not working in the problem. I am given the final Velocity-center of mass, but I need to find the Force in N for a third particle.
 
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If you would give us a complete statement of the problem, we may be able to help you. What you have given us is insufficient to understand what you are doing or what you want.
 
Here is the simple problem.

All of the objects start at rest and move for 32 sec. If the velocity of the center of mass after 32 sec is 6 m/s at an angle of 124, determine the magnitude and direction of mass 3.

m1---4kg----F1--1.8N @ 48
m2---5kg----F2--1.5N @ 245
m3---6kg----F3--?N @ ?

The answer I got was 2.671N @ 143.349

The answer in the book is 3.18N @ 132.4

I used vf=vi+at to find a...

then used Ma_cm= F1+F2+F3 broken into the x and y components
 
It sounds like you were on the right track. Perhaps your mistake was just in the evaluation?

The book's answer seems to have too many significant digits, considering that you were only given forces with two significant digits of accuracy.

You can use Newton's second law for components to get two equations with two variables,

F_{net x} = F_{1x} + F_{2x} + F_{3x} = M_{net} a_x
F_{net y} = F_{1y} + F_{2y} + F_{3y} = M_{net} a_y

Where

F_{3x} = F_3 \cos \theta

and

F_{3y} = F_3 \sin \theta

It would probably be easiest to solve for F3. Substitute that into the other equation, and you have something you can rearrange and evaluate, using the fact that

v_x = a_x t

and

v_y = a_y t
 
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Do those all show up as "LaTeX graphic is being generated, please wait a moment, then reload the page" to anyone else? It looks fine when I preview it, but not when it's actually posted.
 
Zorodius said:
Do those all show up as "LaTeX graphic is being generated, please wait a moment, then reload the page" to anyone else? It looks fine when I preview it, but not when it's actually posted.

It appears the same for me.
 
Thanks for the help. After looking at the problem again, I realized I used cos in both the x and y components. Silly stupid mistake. got the right answer now though.
 

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