Calculating the Magnitude of a Projected Force

[Robb]

Homework Statement

Determine the magnitude of the projection of force F = 700 N along the u axis

Homework Equations

The Attempt at a Solution

A(-2, 4, 4)
r(AO) = 2i -4j - 4k
r(AO mag)= 6
u(AO) = 1/3i - 2/3j - 2/3k
F(AO) = 233.3333i - 466.6667j - 466.6667k

I'm not sure where to go from here. The only info I can figure about u is the 30 degree angle between it and the y axis.

 

[gneill]

This is a vector problem. Do you know how to find the projection of one vector along the direction of another?

 

[Robb]

F dotted with u(AO)

 

[gneill]

F dotted with u(AO)

Hmm. Don't you mean F dotted with u? Note that F lies along the direction OA. So you'll need to construct a vector for F and another for the u unit vector. That 30° angle should come in handy for the latter.

 

[Robb]

F = 233.3333i - 466.6667j - 466.6667k

I guess I'm just not seeing the u unit vector. And, yes, that is what i meant.

 

[gneill]

F = 233.3333i - 466.6667j - 466.6667k

I guess I'm just not seeing the u unit vector. And, yes, that is what i meant.

I think you need to check the signs of your F components. It looks to me from the diagram that F is pointing towards the negative x-axis, positive y-axis, and positive z-axis.

The u unit vector should lie along the u-axis. Use a bit of trig to find the x and y components. It appears to lie in the x-y plane...

 

[Robb]

I think you need to check the signs of your F components. It looks to me from the diagram that F is pointing towards the negative x-axis, positive y-axis, and positive z-axis.

The u unit vector should lie along the u-axis. Use a bit of trig to find the x and y components. It appears to lie in the x-y plane...

u(x) = 4/cos30 = 4.62
u(y) = sqrt(4.62^2 - 4^2) = 2.31
u = .5i +.866j
F(u) = (-233.3333i)(.5) + (466.6667)(.866) = 287.5

They want the answer to two significant figures

 

[gneill]

The simplest approach to finding a unit vector is to imagine that it sits within a unit circle, and then the sine and cosine of a suitable angle will give your the components directly. The magnitude of any vector with unit length is always unity, just like the radius of the unit circle.

The axis u is indicated as having an angle of 30° with respect to the y-axis. So you should expect the y-component to be the cosine of the angle and the x-component to be the sine of the angle. You seem to have found the opposite (and I don't really understand where your value of 4 came from... there's nothing indicated in the figure that ascribes a dimension of 4 in relation to the u-axis).

This is what you have to work with:

 

[Robb]

Yep, I understand that. I was grabbing at straws because I'm not sure how to get values for u.

 

[gneill]

Yep, I understand that. I was grabbing at straws because I'm not sure how to get values for u.

Okay, so are you good to go now?

 

[Robb]

Okay, so are you good to go now?

So, I have u(y) = ucos30 and u(x) = usin30
I guess where I'm lost in all of this is what is u? I need to use the dot product to solve but without a value for the above components how can I do that. Feeling a little blind here.

 

[gneill]

So, I have u(y) = ucos30 and u(x) = usin30
I guess where I'm lost in all of this is what is u? I need to use the dot product to solve but without a value for the above components how can I do that. Feeling a little blind here.

u is a unit vector along the positive u-axis. Refer to the diagram that I posted.

All you need in order to find a projection along a given direction is to dot a given vector with a unit vector in the desired direction. Works with any unit vector; You can try it with your F vector and the unit vectors for the x,y, and z axis if you like. Take the dot product of F with any of the axes unit vectors and you should "extract" that component from the F vector.

 

[Robb]

usin30 + ucos30 = 1
u = .7214
u(x) = .3607
u(y) = .6248
F dot u = 209
To two sig figs F(u) = 210

 

[gneill]

usin30 + ucos30 = 1

No, how do vector components add?

u = .7214
u(x) = .3607
u(y) = .6248

u can't be both a scalar and a vector. Besides, the sine and cosines all by themselves satisfy the requirement of unit vector components. $sin^2 + cos^2 = 1$. So just use the sine and cosine as the components of u.

F dot u = 209
To two sig figs F(u) = 210

You'll need to re-do that with the fixed u.

 

[Robb]

yeah, that wasn't real smart.

F dot u = (-233.3333)(.5) + (466.6667)(.8661) = 287.5

two sig figs = 290

 

[gneill]

yeah, that wasn't real smart.

F dot u = (-233.3333)(.5) + (466.6667)(.8661) = 287.5

two sig figs = 290

You've mixed up the components of u again. Look at the diagram in post #8. Is the sine of the angle along the y-axis or the x-axis?

 

[Robb]

You've mixed up the components of u again. Look at the diagram in post #8. Is the sine of the angle along the y-axis or the x-axis?

sin=x
cos=y

 

[gneill]

sin=x
cos=y

Right. And sin(30°) = 1/2, while cos(30°) = √3 / 2.

 

[Robb]

Right. And sin(30°) = 1/2, while cos(30°) = √3 / 2.

I agree: (-233.333)(.5) + (466.6667)(.8661) = 287.5 or 290

 

[gneill]

I agree: (-233.333)(.5) + (466.6667)(.8661) = 287.5 or 290

D'oh! I apologize. I misread and jumped without engaging my brain

You have indeed got u sorted out now. So your result is correct

 

[Robb]

No problem. My brain has not been engaged this entire problem. I kept thinking I needed to find a value for u when u is simply 1 because it is a unit vector! Thanks for your help, I learned a couple things on this one!