TWO EQUAL FORCES HAVE A RESULTANT EQUAL TO 1 1/2 TIMES THE EITHER FORCE . AT WHAT ANGLE THEY ARE INCLINED TO EACH OTHER
To find the angle at which two equal forces are inclined to each other, given that their resultant is equal to 1 1/2 times either force, we can use the concept of vector addition.
Let's assume that each force has a magnitude of "F" units. Since the forces are equal, the magnitude of each force is the same.
We also know that the resultant of these two forces is 1 1/2 times either force. Therefore, the magnitude of the resultant can be written as (1 1/2) F or (3/2) F.
Now, let's say that the angle between the two forces is "θ".
To calculate the resultant of the forces, we use the formula:
Resultant = √((F^2) + (F^2) + (2 * F * F * cosθ))
Since the forces are equal, this equation simplifies to:
(3/2) F = √(4F^2 * cos^2θ)
Now, let's solve for cosθ:
(3/2) = 2 * cos^2θ
cos^2θ = 3/4
Taking the square root of both sides, we have:
cosθ = √(3/4)
cosθ = √3/2
Now, we need to find the angle "θ" that corresponds to the value of cosθ. Recall that cosθ represents the adjacent side divided by the hypotenuse in a right-angled triangle. The value √3/2 corresponds to a 30-degree angle (θ) in the first quadrant.
Since the two forces are equal, they are inclined to each other at an angle of 30 degrees.
two equal force have their resultant equal to eith
Draw parallelogram with all sides of length one.
cos(T/2) = (1.5/2) = .75
T/2 = 41.4 deg
T = 82.8 degrees
cos T/2=1.5/2 = .75
T/2 = 41.4 deg
T= 41.4*2
= 82.9 •