two force acting at right angle to each other and have a resultant of 98n calculate the magnitude of the force if one of tham make and angle 60 with the resultant
find Fx which is 60 deg from result
Fx^2 + Fy^2 = 98^2
tan 60 = Fy/Fx
so Fy = Fx* tan 60
Fx^2 + Fx^2 tan^2 60 = 98^2
Fx^2 (1+tan^2 60) = 98^2
Fx = 98/sqrt (1+tan^2 60)
To solve this problem, you can use the concept of vector addition and trigonometry.
Let's assume that the magnitude of one force is F1 and the angle it makes with the resultant is 60 degrees. The other force, which is perpendicular to the first force, can be denoted as F2.
The resultant force, which is the sum of the two forces, can be calculated using the Pythagorean theorem:
Resultant force (R) = sqrt((F1^2) + (F2^2))
Given that the magnitude of the resultant force is 98 N, we can substitute this value into the equation and solve for F1:
98 = sqrt((F1^2) + (F2^2))
Now, let's calculate the value of F2. Since the two forces are perpendicular to each other, their vector sum forms a right-angled triangle. Therefore, we can use trigonometry to find F2.
In a right-angled triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. In this case, F2 is opposite the angle of 60 degrees, and R is the hypotenuse.
So, we have the following relation:
sin(60) = F2 / R
Rearranging the equation, we find:
F2 = R * sin(60)
Substituting the given values, we get:
F2 = 98 * sin(60)
Now, to find F1, we can use the relation:
F1 = sqrt((R^2) - (F2^2))
Substituting the values, we get:
F1 = sqrt((98^2) - (98 * sin(60))^2)
Calculating the values:
F2 = 98 * sin(60) ≈ 84.85 N (rounded to two decimal places)
F1 = sqrt((98^2) - (98 * sin(60))^2) ≈ 47.87 N (rounded to two decimal places)
Therefore, the magnitude of the force making an angle of 60 degrees with the resultant is approximately 47.87 N.