Where Did I Go Wrong in Calculating Tension and Normal Force?

[DyslexicHobo]

Homework Statement

Each box weighs 40 lbs, the angles are measured relative to the horizontal. The surfaces are smooth (no friction). Determine the tension in rope A and the normal force exerted by box B by the incline.

Homework Equations

W = mg
ΣF_x = 0
ΣF_y = 0

The Attempt at a Solution

I found the tension of rope C and then normal force exerted on box D by setting up a system of equations.

Solved, the tension equals 15.1 lb, and the normal force exerted is 33.1 lb.

Now I'm having trouble determining the normal force of B and the tension of rope A.

Attempting the solution:
ΣF_x = 0 = 15.1*cos(45) + N*sin(70) - T_a*cos(70)
ΣF_y = 0 = -15.1*sin(45) + N*cos(70) - T_a*sin(70) - 40

Solved using systems of equations, we found the normal force to be .0015lb (unreasonable), and the tension of A to be 31.2 lb (also unreasonable?).

I can't figure out where I went wrong to get this bad answer. Anyone spot some mistakes? Thanks.

 

[edziura]

Is there supposed to be a picture attached?

 

[DyslexicHobo]

Haha, yes, there is. I completely forgot to attach it. Why can't I edit my original post?

Here's the pic (see attached)

 

[Hootenanny]

Why can't I edit my original post?

You are only permitted to edit your posts for a maximum of one hour after they were originally posted.

 

[alphysicist]

Hi DyslexicHobo,

The Attempt at a Solution

I found the tension of rope C and then normal force exerted on box D by setting up a system of equations.

Solved, the tension equals 15.1 lb, and the normal force exerted is 33.1 lb.

Now I'm having trouble determining the normal force of B and the tension of rope A.

Attempting the solution:
ΣF_x = 0 = 15.1*cos(45) + N*sin(70) - T_a*cos(70)
ΣF_y = 0 = -15.1*sin(45) + N*cos(70) - T_a*sin(70) - 40

The tension of rope A is upwards and so I believe the corresponding term in the y equation needs to be positive to match the other terms.