Finding Resultant Force using Component Method

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SUMMARY

The discussion focuses on calculating the resultant force using the component method for four forces: F1 = 300g at 30°, F2 = 450g at 110°, F3 = 400g at 230°, and F4 = 270g at 298°. The user successfully converted the forces into Newtons and expressed them in unit vector notation. The resultant force vector was calculated as R = (-1.48)i + (2.61)j with a magnitude of 2.99 N. The direction of the resultant can be determined from its components, and it is essential to include units in the final results.

PREREQUISITES
  • Understanding of vector notation and unit vectors
  • Knowledge of trigonometric functions (sine and cosine)
  • Familiarity with Newton's second law (F=ma)
  • Ability to perform vector addition in component form
NEXT STEPS
  • Learn how to calculate the angle of a resultant vector using arctangent
  • Study the concept of equilibrium in force systems
  • Explore the use of free-body diagrams in physics
  • Investigate advanced vector operations, such as cross products
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying mechanics, as well as educators and anyone involved in laboratory experiments related to force analysis and vector calculations.

jeff12
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Homework Statement



This is the force table lab that we are doing.

F _1 = 300g @ 30^o
F _2 = 450g @ 110^o
F _3 = 400g @ 230^o
F _4 = 270g @ 298^o

Finding resultant force using Component Method

1. Express each force F1, F2, and F3 in unit vector notation. Take the origin to be at the center of the force table (at pivot point) with the +x axis along 0^o and +y-axis along 90^o.

2. Use the component method to obtain the resultant force vector Fcomp in unit vector notation. Calculate the magnitude and direction.

Homework Equations


sinθ=opp/hyp
cosθ=adj/hyp
F=ma

The Attempt at a Solution



His directions are kind of confusing to me.
For the step 1 I found:

I converted the mass into kg and then found the force.

F _1 = 2.94N
F _2 = -4.41N
F _3 = -3.92N

Then I express each force into unit vector notation.

F _1 = 2.94cos30i +2.94sin30j
F _2 = -4.41cos70i +4.41sin70j
F _3 = -3.92cos50i -3.92sin50j

Did I do this part right? Please check it for me.

I don't know where to start for step 2. Do I find F4?
 
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jeff12 said:

Homework Statement



This is the force table lab that we are doing.

F _1 = 300g @ 30^o
F _2 = 450g @ 110^o
F _3 = 400g @ 230^o
F _4 = 270g @ 298^o

Finding resultant force using Component Method

1. Express each force F1, F2, and F3 in unit vector notation. Take the origin to be at the center of the force table (at pivot point) with the +x axis along 0^o and +y-axis along 90^o.

2. Use the component method to obtain the resultant force vector Fcomp in unit vector notation. Calculate the magnitude and direction.

Homework Equations


sinθ=opp/hyp
cosθ=adj/hyp
F=ma

The Attempt at a Solution



His directions are kind of confusing to me.
For the step 1 I found:

I converted the mass into kg and then found the force.

F _1 = 2.94N
F _2 = -4.41N
F _3 = -3.92N
You should leave the signs off of the force magnitudes. Force magnitudes are always positive (absolute) values. It's the information from the angles where you form the components that will turn them into vectors with direction.
Then I express each force into unit vector notation.

F _1 = 2.94cos30i +2.94sin30j
F _2 = -4.41cos70i +4.41sin70j
F _3 = -3.92cos50i -3.92sin50j
Okay, I see you've sorted out how to use the trig functions with the given angles. Well done. At this point you should reduce the vectors to purely numerical component values. That is, evaluate each component, getting rid of the trig.
I don't know where to start for step 2. Do I find F4?
The instructions don't mention F4 at this point, so I'd just combine F1,F2, and F3 into a resultant. Do you no how to add vectors in unit vector form?
 
gneill said:
The instructions don't mention F4 at this point, so I'd just combine F1,F2, and F3 into a resultant. Do you no how to add vectors in unit vector form?

Is the formula:
R=A+B+C which would be R=(F1x+F1y)i+(F2x+F2y)j+(F3x+F3y)k?
 
jeff12 said:
Is the formula:
R=A+B+C which would be R=(F1x+F1y)i+(F2x+F2y)j+(F3x+F3y)k?
No, you only add like with like. That is, sum all the x-components to yield the net x component, sum all the y-components to yield the net y-component,...
 
Okay,
R=(-1.48)i+(2.61)j
Magnitude:
R=2.99
but what is the direction? Is that right?

I just want to clarify the R that we found is considered the F4 but found mathematically right? Because in the lab we were given F1, F2, ,F3 and we had to find F4 using the weights.
 
jeff12 said:
Okay,
R=(-1.48)i+(2.61)j
Magnitude:
R=2.99
but what is the direction? Is that right?
Looks okay. The direction is encoded in the components. Just sketch the resultant using the components and you'll see the direction. You can use these rectangular components to work out the magnitude and the angle if you want the polar version of the vector. Oh, always include units on results! That should be R=[(-1.48)i+(2.61)j] N, and |R| = 2.99 N.
I just want to clarify the R that we found is considered the F4 but found mathematically right?
I don't know, I wasn't in your lab :smile: But it seems a likely scenario that some fourth force was employed to balance the force table setup. Did you have to play with a fourth force to try to find an equilibrium? If so, in order to balance the resultant of the forces F1 through F3 you'd have to apply a force of the same magnitude as the resultant but in the opposite direction (i.e. oppose it equally).
 
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