Specify force magnitude and angle

In summary: Yes, that looks right. The z-component is negative because the vector F1 is directed below the x-y plane.To calculate the angles of F2 I used:cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 72.36° cos y = (F2y/F2) = (650N / 700 N) ---> y = 21.79°cos z = (F2z/F2) = (150N / 700 N) ---> z = 77.63°Yes, now that you've provided the equations and the diagram, I think you're on
  • #1
mtjc1x
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Homework Statement


See picture.

Homework Equations

The Attempt at a Solution


I tried drawing a parallelogram, but it didn't help, so I'm stuck. I'd need an idea in the right direction, if possible.
 

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  • #2
mtjces said:

Homework Statement


See picture.

Homework Equations

The Attempt at a Solution


I've decomposed F1 into F1x F1y and F1z..
F1x = 150 * sqrt 3
F1y = 150
F1z =150 * sqrt 2[

Now I'm stuck../QUOTE]
 
  • #3
I'm not seeing how you've accomplished your decomposition of F1. The magnitude of the vector comprised by those components would be about 367 N, which is larger than the given magnitude of F1. So you might want to re-think your method (whatever that was...).

The diagram is very helpful because it gives you the angles between F1 and each of the coordinate axes. This should make finding the projections of F1 on each of the axes easy to find.
 
  • #4
mtjces said:

Homework Statement


See picture.

Homework Equations

The Attempt at a Solution


I tried drawing a parallelogram, but it didn't help, so I'm stuck. I'd need an idea in the right direction, if possible.
I don't know what sort of parallelogram you drew, but this is a 3-D kinda problem.
 
  • #5
SteamKing said:
I don't know what sort of parallelogram you drew, but this is a 3-D kinda problem.

yes, I was going to draw a parallelepiped to see if it could help me.
 
  • #6
gneill said:
I'm not seeing how you've accomplished your decomposition of F1. The magnitude of the vector comprised by those components would be about 367 N, which is larger than the given magnitude of F1. So you might want to re-think your method (whatever that was...).

The diagram is very helpful because it gives you the angles between F1 and each of the coordinate axes. This should make finding the projections of F1 on each of the axes easy to find.

F1z = cos 60 * 300N = 150N
F1x = cos 45 * 300N = 150*sqrt2 N
F1y = cos 60 * 300N = 150 N

Would this be correct?
I'm confused about the next step..
 
  • #7
mtjces said:
F1z = cos 60 * 300N = 150N
F1x = cos 45 * 300N = 150*sqrt2 N
F1y = cos 60 * 300N = 150 N

Would this be correct?
For the Z component the angle was not 60 degrees. What does the diagram show it to be? Do you expect the Z-compoennt to be less than or greater than zero?

Check your value for cosine 45 degrees. A cosine should not be > 1.*

* EDIT: Oops. I didn't process the fact that the 300 N was divided by two giving 150 N, so the cosine being ##\sqrt{2} / 2## is satisfied! Sorry about that!
I'm confused about the next step..
You'll want to write a vector equation that expresses your desired result. Fix your F1 vector first though.
 
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  • #8
gneill said:
For the Z component the angle was not 60 degrees. What does the diagram show it to be? Do you expect the Z-compoennt to be less than or greater than zero?

Check your value for cosine 45 degrees. A cosine should not be > 1.

You'll want to write a vector equation that expresses your desired result. Fix your F1 vector first though.

F1z = cos 120 * 300N = -0.5 * 300N = -150N (but how does that make sense?)

cos 45 = (√2)/2
so F1x = 300 N * cos 45 = 150 N * √2 = 150√2 N

F1y = cos 60 * 300N = 150 N

Is this right now?
 
  • #9
mtjces said:
F1z = cos 120 * 300N = -0.5 * 300N = -150N (but how does that make sense?)

cos 45 = (√2)/2
so F1x = 300 N * cos 45 = 150 N * √2 = 150√2 N

F1y = cos 60 * 300N = 150 N

Is this right now?
Yes, that looks right. The z-component is negative because the vector F1 is directed below the x-y plane.
 
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  • #10
gneill said:
Yes, that looks right. The z-component is negative because the vector F1 is directed below the x-y plane.

so the idea with F2 would be to cancel out the F1z and F1x and amplify F1y by 650N?
So I need a
F2z = 150 N
F2x = -150√2 N
F2z = 650 N

Am I going in the right direction?
 
  • #11
Yes, but sort out your component subscripts. You've listed F2z twice. It would help if you put them in standard order, too: x, y, z. Otherwise, I think you're there.
 
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  • #12
gneill said:
Yes, but sort out your component subscripts. You've listed F2z twice. It would help if you put them in standard order, too: x, y, z. Otherwise, I think you're there.

F2x = -150√2 N
F2y = 650 N
F2z = 150 N

How do I calculate F2?
 
  • #13
mtjces said:
F2x = -150√2 N
F2y = 650 N
F2z = 150 N

How do I calculate F2?
You don't know how to calculate the magnitude from the three force components?
 
  • #14
mtjces said:
F2x = -150√2 N
F2y = 650 N
F2z = 150 N

How do I calculate F2?

F2 = √((-150√2)^2 + 150^2 + 650^2) = 700 N

To calculate the angles of F2 I used:

cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 72.36°
cos y = (F2y/F2) = (650N / 700 N) ---> y = 21.79°
cos z = (F2z/F2) = (150N / 700 N) ---> z = 77.63°

Correct?
 
  • #15
mtjces said:
F2 = √((-150√2)^2 + 150^2 + 650^2) = 700 N

To calculate the angles of F2 I used:

cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 72.36°
cos y = (F2y/F2) = (650N / 700 N) ---> y = 21.79°
cos z = (F2z/F2) = (150N / 700 N) ---> z = 77.63°

Correct?
Sorry, messed up x:
cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 107.64
 

1. What is a force magnitude?

The magnitude of a force is the measure of its strength or intensity. It is typically measured in units of Newtons (N) in the metric system or pounds (lbs) in the imperial system.

2. How do you calculate the magnitude of a force?

The magnitude of a force can be calculated by multiplying the mass of an object by its acceleration, using the equation F=ma. Alternatively, it can also be calculated using trigonometric functions if the force is acting at an angle.

3. What is an angle in the context of force?

An angle in the context of force refers to the direction in which the force is acting. It is typically measured in degrees or radians and can be used to determine the direction and orientation of the force.

4. How do you specify the angle of a force?

The angle of a force can be specified using a reference point or axis, such as the x-axis or y-axis. It can also be specified using the direction of motion or the location of the force relative to an object.

5. Why is it important to specify both the magnitude and angle of a force?

Specifying both the magnitude and angle of a force is important because it provides a complete description of the force and its effects. It allows us to accurately calculate the resulting motion or changes in an object, and helps us understand the forces at play in a given system.

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