SOLUTION: B-6 Machine Trades On a crankshaft, AB=4.2 in., and the connecting rod AC= 12.5 in. Calculate the size of angle A when the angle at C is 12 degrees. I know the answers are 26 degre

Question 112952This question is from textbook Mathematics for the Trades
: B-6 Machine Trades On a crankshaft, AB=4.2 in., and the connecting rod AC= 12.5 in. Calculate the size of angle A when the angle at C is 12 degrees. I know the answers are 26 degrees, and 130 degrees, but cannot come up with how they got this. HELP and a Very Big THANKS. This question is from textbook Mathematics for the Trades

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
I'm not sure that I entirely understand the geometry of this problem, but let me give you
some thoughts and maybe you can straighten me out or maybe get enough of a clue from the
following explanation to figure it out.
.
I interpret your description as follows. Point B is on the main axis of the crankshaft
and point A is on the crank. The throw radius of the crankshaft is 4.2 inches. One end
of the connecting rod connects to the crank at point A. The other end fastens to the piston
pin at point C and the length of the connecting rod is 12.5 inches.
.
Therefore, you have a triangle that is formed by points B, A, and C ...
.
The 12 degree angle at point C is formed by the connecting rod and the line that that runs
from the piston pin (point C) to the main axis of the crankshaft (point B).
.
Let's look at this triangle BAC. Side BA is opposite angle C and is 4.2 inches long. Side AC
is opposite of angle B and is 12.5 inches long. And side BC is opposite of angle A and is of
unknown length. I am assuming that the crank is approximately in the 11 o'clock position relative to
its spin about the main axis of the crankshaft ... that is to say that point A is approximately
at the 11 o'clock position relative to the axis of its revolution at point B.
.
We are trying to find angle A ... the angle formed by the crank arm and the connecting rod.
.
We can apply the law of sines which says that in triangle ABC there is a relationship:
.

.
where the following definitions apply ...
.
BC is the length of the side opposite angle A which leads to the relationship
.
BA is the length of the side opposite angle C which leads to the relationship
.
AC is the length of the side opposite angle B which leads to the relationship
.
From the description above and the problem we have BA = 4.2 inches and angle C is 12 degrees.
Therefore, we can substitute for BA and for C to get the ratio .
.
Next we know that the connecting rod (side AC) has a length 12.5 inches, but angle B
which is opposite side AC is unknown. Therefore, the relationship here is and
by substituting for AC this relationship becomes . The law of sines says
that this relationship has to be equal to the previous ratio we established. That is:
.

.
As usual, you can solve this proportion by cross multiplying ... multiply 4.2 times sinB and
set that product equal to the product of 12.5 times sin(12). That results in the equation:
.

.
A calculator will tell you that sin(12) = 0.20791169 and when you multiply that by 12.5
you get 2.598896135. This reduces the equation to:
.

.
Divide both sides of this equation by 4.2 and you end up with:
.

.
We can now solve for angle B by taking arcsine 0.618784794 to get that angle B is 38.22744791 degrees.
.
Now we have that angle C is 12 degrees and angle B is 38.22744791 degrees. The missing
angle is angle A. But we can find angle A by recognizing that the sum of the three angles
in the triangle must be 180 degrees. Therefore:
.
180 = 12 + 38.22744791 + angle A
.
Subtracting 12 + 38.22744791 from both sides results in:
.
180 - 12 - 38.22744791 = angle A
.
and this reduces to:
.
129.7725521 degrees = angle A
.
This is pretty close to the 130 degree answer that you gave.
.
Next I would suspect that the crank (point A) is at the 5 o'clock position relative to
point B.
.
I haven't figured the geometry for this, but note that 38.22744791 degrees we got previously
minus the 12 degrees would give an answer of 26.22744791 degrees. I wonder ...?
.
Hope this turns you on to something. Sorry about the complex discussion above, but it's
hard to do a geometry discussion in this forum where the drawing support is limited.

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