Calculate Fb for 3 Forces on a Ring with Mass 100kg and Accelerating .5m/s2

[zaper]

Three forces are applied to a ring (as shown in the photo) that lies on a frictionless surface in the xy plane. The ring has a mass of 100 kg. Fa=200N, Fc=240N and the angle between Fa and Fb is 135°.

What is Fb if:

The system is stationary?

The system accelerates at .5 m/s2?

For some reason I just can't get a grasp on this problem. I understand that to be stationary all forces must cancel out, but I can't figure out how with Fc having a stronger pull in the x direction than Fa how Fb (which appears to go straight down) can stop Fc.

 

[tiny-tim]

hi zaper!

Three forces are applied to a ring (as shown in the photo) that lies on a frictionless surface in the xy plane. The ring has a mass of 100 kg. Fa=200N, Fc=240N and the angle between Fa and Fb is 135°.
…
… I can't figure out how with Fc having a stronger pull in the x direction than Fa how Fb (which appears to go straight down) can stop Fc.

you aren't told the angle between F_c and the other two …

so you can make it anything you like!

 

[zaper]

So Basically to not move I should make the angle between Fb and Fc 135 as well which means that Fa and Fc will cancel x-wise and Fb will have to be 2*Fa*cos(45)?

 

[tiny-tim]

no, because F_a is 200 N and F_c is 240 N

 

[zaper]

Oh yeah. So then 200cos(45)=240cos(x) so Fc is at 53.9. This means Fb=200sin(45)+240sin(53.9)which is 335.3

 

[tiny-tim]

no, not 45°, you need to use an unknown θ

 

[zaper]

Yeah I edited my previous post so hopefully it's correct. I'm sorry. My brain is not working this morning

 

[tiny-tim]

ahh!

Oh yeah. So then 200cos(45)=240cos(x) so Fc is at 53.9. This means Fb=200sin(45)+240sin(53.9)which is 335.3

yes! (i haven't checked the figures, but …) that method looks fine

 

[zaper]

Ok so that solves the first part. Now for the second part since it goes .5m/s2 in the x direction and the ring is 100 kg then Fc is 50 N greater than Fa in the x direction so 200cos(45)=240cos(x)-50?

 

[zaper]

If this method is correct then I get that Fc is at 37.1. This means that Fb=200sin(45)+240sin(37.1) which is 286.2

 

[tiny-tim]

Ok so that solves the first part. Now for the second part since it goes .5m/s2 in the x direction and the ring is 100 kg then Fc is 50 N greater than Fa in the x direction so 200cos(45)=240cos(x)-50?

yes, but you'll need a y equation also

 

[zaper]

I have that in my last post I believe

 

[tiny-tim]

ahh!

If this method is correct then I get that Fc is at 37.1. This means that Fb=200sin(45)+240sin(37.1) which is 286.2

yes, that looks fine too