- #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
Click For Summary
Homework Help Overview
The discussion revolves around calculating the net force from two forces acting at an angle, involving vector representation and the cosine law. Participants are exploring the relationships between the forces and their angles in a two-dimensional context.
Discussion Character
- Mixed
Approaches and Questions Raised
- Participants are attempting to apply the cosine law to find the resultant force and are discussing the angles involved in the problem. There are questions about the correct angles to use and how to resolve the forces into components.
Discussion Status
There is ongoing exploration of the angles and methods to calculate the net force. Some participants have provided guidance on using vector representation and resolving components, while others are clarifying misunderstandings about the angles involved.
Contextual Notes
Participants are required to follow a specific method outlined in their lesson, which emphasizes the use of the cosine law, despite some suggesting alternative approaches. There is confusion regarding the angles derived from the diagram and how they relate to the forces being analyzed.
- #32
The Vinh
- 46
- 1
Oh and you have a really " cool " teacher
- #33
- 42,842
- 10,509
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
- 42,842
- 10,509
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
- 42,842
- 10,509
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
- 42,842
- 10,509
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
- 42,842
- 10,509
Ok!alexandria said:
- #42
alexandria
- 168
- 2
thank you so much for your help, i really appreciate it :)
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