if the sum of the forces in the x-direction = 0 = -0.25w(cos(x)) + w(cos(y)) and the sum of the forces in the y-direction = 0 = 0.25w(sin(x)) + w(sin(y)) - w

Question

One right triangle may have sides of  m = 0.25w(cos(x))  and  h = 0.25w(sin(x)) also

a second right triangle  may have sides of n = w(cos(y))  and  h = w(sin(y)) also if the

law of cosines is used m + n = c and the two other sides are a = 0.25w and b = w

find the angle x in degrees?

Answer 1

From the 1st eqn.

-0.25cos(x) + cos(y) = 0

cos(x) = 4cos(y) --------------------- (1)

From the 2nd eqn,

(1/4)sin(x) + sin(y) - 1 = 0

sin(y) = 1 - (1/4)sin(x)

sin^2(y) = (1 - (1/4)sin(x))^2

1 - cos^2(y) = (1 - (1/4)sin(x))^2

Using (1),

1 - (1/16)cos^2(x) = (1 - (1/4)sin(x))^2

1 - (1/16)(1 - sin^2(x)) = (1 - (1/4)sin(x))^2

1 - 1/16 +(1/16)sin^2(x) = 1 - (1/2)sin(x) + (1/16)sin^2(x)

15/16 = 1 - (1/2)sin(x)

(1/2)sin(x) = 1/16

sin(x) = 1/8 = 0.125

x = sin^(-1)(0.125)

x = 7.181 degrees

 

Answer 2

This problem appears to be a vector problem in which three forces are interacting, with one force W1 applied at an angle x to the horizontal, another W2 at an angle y to the horizontal, and a third horizontal force W3. When the forces are resolved into their horizontal and vertical components we have:

-0.25wcos(x)+wcos(y) ("eastwards") and 0.25wsin(x)+wsin(y)-w ("northwards") respectively.

Diagrammatically if we draw horizontal and vertical axes, W1, with a magnitude 0.25w, pulls in a negative direction, sloping backwards (\) at angle x to the horizontal; W2, magnitude w, pulls in a positive direction sloping forwards (/) at angle y to the horizontal; W3, magnitude w, pulls horizontally "westwards".

If the forces cancel out as indicated, 0.25wcos(x)=wcos(y); so 0.25cos(x)-cos(y)=0; and 0.25sin(x)+sin(y)=1 (because w's cancel out). 

So 0.25cos(x)=cos(y); squaring both sides: 0.0625(cos(x))^2=(cos(y))^2 and, replacing cos^2 with 1-sin^2, we get  0.0625(1-(sin(x))^2)=1-(sin(y))^2. This reduces to (sin(y))^2-0.0625(sin(x))^2=0.9375.

This allows us to build an equation in sin(x) only, by substituting sin(y)=1-0.25sin(x): 

(1-0.25sin(x))^2-0.0625(sin(x))^2=0.9375.

1-0.5sin(x)+0.0625(sin(x))^2-0.0625(sin(x))^2=0.9375; 0.5sin(x)=0.0625; sin(x)=0.125 and x=7.18 degrees (approx). x could also be 180-7.18=172.82, so better describing the direction of W1.

[The question doesn't ask for y, but sin(y)=1-0.25sin(x)=1-0.25*0.125=0.03125, making y=1.79 degrees approximately.]

The discourse in the text of the question seems slightly puzzling, but the triangle involving m and h seems to be a representation of W1 represented by the hypotenuse of the triangle. The second triangle similarly represents W2, so we have W1 and W2 with their respective angles. Since we have two angles, we can find the third: 180-(7.81+1.79)=170.4, where the three angles lie on a line. Perhaps this angle is the angle actually required? The triangle so formed would give us the resultant force, W, of W1 and W2 using the cosine rule: W^2=W1^2+W2^2-2W1W2cos170.4=1.5555w^2, making W=1.247w.