Finding F1 - Summing forces on a box

[jdawg]

Finding F1 -- Summing forces on a box

Homework Statement

There are two forces on the 2.80 kg box in the overhead view of the figure but only one is shown. For F1 = 10.9 N, a = 10.2 m/s2, and θ = 34.9°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)

I'm not sure if this link will work, but here's the picture: http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c05/fig05_32.gif

Homework Equations

∑F=ma
∑F=F₁+F₂...+

The Attempt at a Solution

Ok so I've been stuck on this one for a while. I started out finding the sum of the forces:
∑F=(2.8kg)(10.2m/s²)=28.56 N

Then I tried solving for F₂:
28.56N=10.9N+F₂
F₂=17.66

Then I found the x and y components of the acceleration:
a_x=10.2*sin(34.9)=5.84m/s²
a_y=10.2*cos(34.9)=8.37m/s²

Could you multiply a_x and a_y by the mass and get the F_1x and F_1y?
I'm so confused, could someone please help guide me through this problem?

 

[rock.freak667]

The x and y components of acceleration are the net accelerations in those directions. Multiplying those by the mass will give you the net forces in the x and y direction.

If you sum the forces in the x direction, they should add up to the resultant force in the x-direction. A similar exercise will yield the resultant force in the y-direction.

You need to use this fact to help you find F2.

 

[jdawg]

Ok, so I should multiply a_x*mass to get ∑F_x? which would give me
∑F_x=16.35
And for the y component: ∑F_y=23.44

So then would you find F₁? To me it looks like F₁ only has an x component... So in unit vector notation: 10.9i+0j?

And then could you solve for F₂? Or am I still going about this the wrong way?

 

[jackarms]

Your approach to the problem is just a bit off (at least in your first post -- in your second you're working through it better). Remember that Newton II is a vector equation, so you can't use magnitudes for the force or acceleration directly. You can only use magnitudes if you do the equation by component. Set up the equation like this:
\Sigma \stackrel{\rightarrow}{F} = m \cdot (\stackrel{\rightarrow}{F}_{1} + \stackrel{\rightarrow}{F}_{2})

Then split that up into components to solve for F_{2x} and F_{2y}.

 

[jdawg]

I'm sorry, I still don't understand. I tried plugging in values into the formula:
28.56=2.8(10.9+F₂)
F₂=-0.7?
Is the acceleration vector the same as the F₂ vector, or would the acceleration vector be along the ∑F vector?

 

[jackarms]

No, for the first part, you're using the wrong values. In the problem statement, you're only given magnitudes and directions for F1 and for acceleration. Whenever you add vectors (like adding F1 and F2), you can't just add magnitudes together, because it's not arithmetic addition, but instead vector addition, so it has to be done by x and y components of the two adding vectors.

And acceleration will point in the same direction as the net force.

 

[jackarms]

Sorry, I put in my equation wrong in my earlier post. It should actually be:

\Sigma \stackrel{\rightarrow}{F} = (\stackrel{\rightarrow}{F}_{1} + \stackrel{\rightarrow}{F}_{2}) = m \cdot \stackrel{\rightarrow}{a}

 

[jdawg]

It almost makes sense! So would my x and y components of the acceleration vector be the same as my x and y components of the ∑F vector?

∑F_x=5.84N
∑F_y=8.37N

Ohh that's ok, that formula makes a whole lot more sense now!

So can you find the x and y components of F₁ vector by just looking at the picture? It doesn't have a y component, right? Isn't it just moving in the x direction? So: 10.9i+0j?

 

[jackarms]

Just about, but not quite. The acceleration vector is parallel to net force, but that doesn't mean their the same. Force isn't the same thing as acceleration, right? Look at the equation for Newton II -- net force is a scaled copy of acceleration, so to go from acceleration to net force, you have to multiply by mass. Also, when you found your acceleration components, you got the right magnitudes, but not direction. Both your x and y components are positive, but in the diagram acceleration points in the third quadrant, so both x and y should be negative. Always look at your diagrams to check signs.

And yes, you can assume the angle between F1 and the x-axis is zero, so the x component is the magnitude, and it doesn't have a y component.

 

[jdawg]

Ohh, I can't believe I messed up on the directions of the acceleration vector components!
Sorry, I'm a little thick headed. Are you saying that I can multiply the acceleration components by the mass and get the components of the net force?
So: ∑F_x=16.35
∑F_y=23.44

 

[jackarms]

Yes, that's just a result of the equation \Sigma F = m \cdot a (but remember to check the signs on net force too.) Then, just resolve F1 into its components (as you did), and then you can solve for the components of F2, and then get stuff like the direction and magnitude of it.

 

[jdawg]

Ok yay! I got part A. To find the magnitude I tried to just do ∑F=F₁+F₂
28.5=10.9+F₂
F₂=17.66
But I'm pretty sure that's wrong...

 

[jackarms]

It's the same mistake again -- that equation can't be used to solve for a magnitude, because it's vector addition. Try instead finding the magnitude using the components.

 

[jdawg]

Oh yeah, sorry! Would you just combine the x components?