Engr Mechanics Statics Problem (summation of forces)

[savcicEX]

Homework Statement

The problem is the following;
Three smooth homogeneous cylinders A, B, and C are stacked in a V-shaped trough. Each cylinder has a diameter of 500 mm and a mass of 100kg. Determine the forces exerted on cylinder A by the inclined surfaces.

Homework Equations

This problem requires you to sum the forces in the X and the Y direction and solve for the 2 normal forces. my problem is that I end up with an equation with 3 variables. A diagram drawn on paint is attached.

The Attempt at a Solution

I first tried to draw a FDB of cylinder A exclusively (Fb, Fc, N1 & N2) but ended up with an equation with 3 variables
0.366Fb - 1.366 Fc +1.366N1=981N (my N1 is on the left side of cylinder A. N2 canceled out).
Then I tried to study cyl. B to later equal the forces but got 0=-566.38 so I gave up that route.
A problem with the way I tried to solve this problem is that I didnt take into consideration the diameter at all.

How would you go on solving this problem. btw the answers are N1=1304 N, N2=829 N

 

[Steve4Physics]

This question is 14+ years old at the time of answering. Since there is little chance of communications with the OP I’ve provided a fairly complete answer.

The (badly drawn) diagram in Post #1 is unviewable for me – it's just blank. It is a bmp (bitmap) file. If downloaded, it can be viewed using other software (I used LibreOfiice Draw). In case anyone else has the same problem I've reproduced it here as a jpg:

The normal force exerted by B on A is simply equal to the downhill component of B’s weight: mgsinθ = 100*9.81sin(30º) = 490.5N. This has:
x-component = 490.5cos((30º) = 424.79N (to the right)
y-component = 490.5sin((30º) = -245.25N (down)

The normal force exerted by C on A is the downhill component of C’s weight: mgsinθ = 100*9.81sin(45º) = 693.67N. This has:
x-component = 693.67cos(45º) = -490.5N (to left)
y-component = 693.67sin(45º) = -490.5N (down)

A free body diagram for A is needed really, but I’ll do without it...

Left N₁ and N₂ be the normal forces exerted by the left and right sides of the trough on A.

Resolving forces on A horizontally:
N₁ sin(30º) – N₂ sin(45º) + 424.79 – 490.5 = 0
0.5N₁ – 0.7071N₂ = 65.71 (equation 1)

Resolving forces on A vertically:
N₁ cos(30º) + N₂cos((45º) -245.25 – 490.5 – 981 = 0
0.866 N₁ + 0.7071 N₂ = 1716.75 (equation 2)

Solving equations 1 and 2 gives N₁ = 1305N and N₂ = 830N, matching the ‘official answers (allowing for rounding error).

Edit: minor changes only.