- #31
The Vinh
- 46
- 1
Oh, ok, so if you must use it then magnitude and angle are easy to find but the real problem is the direction of net force.alexandria said:
Forces - calculating the net force
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SUMMARY
The discussion focuses on calculating the net force from two vectors: 32.0 N in the +z direction and 38.0 N in the x-y plane. The correct angle between the two forces is determined to be 125 degrees, leading to a resultant force calculated using the cosine law. The final net force is approximately 32.9 N at an angle of 72 degrees North of East. The participants clarify the use of the sine and cosine laws to resolve the forces accurately.
PREREQUISITES- Understanding of vector components in physics
- Familiarity with the cosine law for calculating resultant forces
- Knowledge of trigonometric functions, specifically sine and cosine
- Ability to interpret force diagrams and angles
- Study the cosine law in detail, particularly its application in physics problems
- Learn how to resolve vectors into their components using trigonometric functions
- Practice drawing and interpreting force diagrams for complex systems
- Explore the concept of resultant forces and their directions in vector addition
Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators teaching force resolution methods.
- #32
The Vinh
- 46
- 1
Oh and you have a really " cool " teacher
- #33
- 42,802
- 10,495
You quote the sine formula, but you did not say which angle in the picture is the A in the formula.alexandria said:
It needs to be opposite the side you divide the sine by.
Which angle do you need to find to answer the question?
- #34
alexandria
- 168
- 2
@haruspex
This was the previous diagram i had made, it forms a triangle instead of a parallelogram. I labelled angle A .
SinA / 38.0 N = Sin55° /32.9 N
This was the previous diagram i had made, it forms a triangle instead of a parallelogram. I labelled angle A .
SinA / 38.0 N = Sin55° /32.9 N
- #35
- 42,802
- 10,495
Yes, that looks good.alexandria said:
- #36
alexandria
- 168
- 2
Ok so here is my final answer38.
c = (a^2 + b^2 – 2abcosC) ½
c = ([32.0 N]^2 + [38.0 N]^2 - 2[32.0 N] x [38.0 N]cos55°) ½
c = 32.9 NSinA/a = SinB/b = SinC/c
SinA / 38.0 N = Sin55° /32.9 N
A = 72 degrees (approx.)
net force = 32.9 N (North 72 degrees East)
does this look right?
c = (a^2 + b^2 – 2abcosC) ½
c = ([32.0 N]^2 + [38.0 N]^2 - 2[32.0 N] x [38.0 N]cos55°) ½
c = 32.9 NSinA/a = SinB/b = SinC/c
SinA / 38.0 N = Sin55° /32.9 N
A = 72 degrees (approx.)
net force = 32.9 N (North 72 degrees East)
does this look right?
- #37
- 42,802
- 10,495
Draw that direction on your diagram. Does it look right?alexandria said:
- #38
alexandria
- 168
- 2
is the correct direction (East 72 degrees North)
- #39
- 42,802
- 10,495
I didn't say your answer was wrong, just asked if it looked reasonable. You should always check that where possible.alexandria said:
- #40
alexandria
- 168
- 2
oh, ok, so i based the direction of this diagram.
the force is pulling the object 72 degrees east of north
the force is pulling the object 72 degrees east of north
- #41
- 42,802
- 10,495
Ok!alexandria said:
- #42
alexandria
- 168
- 2
thank you so much for your help, i really appreciate it :)
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