Two forces ,whose resultant is 100N are perpendicular to each other .if one of them makes an angle of 60degrees with the resultant calculate its magnitude
draw the diagram. You have a nice 30-60-90 right triangle with a diagonal of 100, so calculating the sides is easy.
right?
50N
To calculate the magnitude of the force that makes an angle of 60 degrees with the resultant force, we can use the concept of vector addition and trigonometry.
Let's assume the two forces are A and B, with the resultant force being R. We are given that the magnitude of the resultant force R is 100 N.
Since the two forces are perpendicular to each other, we can apply the Pythagorean theorem to find the magnitude of the individual forces A and B.
By Pythagoras' theorem, we have:
R^2 = A^2 + B^2
Since A makes an angle of 60 degrees with R, we can use trigonometry to find the value of A in terms of R.
Using the cosine function, we have:
cos(60) = A / R
A = R * cos(60)
Substituting the value of A in the Pythagorean theorem equation, we get:
R^2 = (R * cos(60))^2 + B^2
Simplifying the equation:
R^2 = R^2 * cos^2(60) + B^2
R^2 - R^2 * cos^2(60) = B^2
R^2(1 - cos^2(60)) = B^2
(R^2 * sin^2(60)) = B^2
By applying the sine function:
sin(60) = √3/2
(R^2 * (3/4)) = B^2
Finally, we can solve for B:
B = √((R^2 * (3/4)))
Substituting the value of R into the equation, we find:
B = √((100^2 * (3/4)))
B = √(10000 * (3/4))
B = √(7500)
B ≈ 86.60 N
Therefore, the magnitude of the force that makes an angle of 60 degrees with the resultant force is approximately 86.60 N.
To calculate the magnitude of one of the forces that makes an angle of 60 degrees with the resultant, we can use vector addition and trigonometry.
Let's denote the magnitude of the first force as F1 and the magnitude of the second force as F2. Since the resultant is 100 N, we can represent it as the vector R = 100 N.
We know that the forces are perpendicular to each other, which means the angle between them is 90 degrees. Therefore, we can use the Pythagorean theorem to determine the relationship between F1, F2, and R:
R^2 = F1^2 + F2^2
Since F1 makes an angle of 60 degrees with the resultant, we can use trigonometry to express F1 in terms of R:
F1 = R * cos(60)
Substituting this into our equation, we get:
R^2 = (R * cos(60))^2 + F2^2
Now, we can solve for F2:
F2^2 = R^2 - (R * cos(60))^2
F2 = sqrt(R^2 - (R * cos(60))^2 )
Plugging in the given value for R:
F2 = sqrt((100 N)^2 - (100 N * cos(60))^2)
F2 = sqrt(10000 N^2 - (100 N * 0.5)^2)
F2 = sqrt(10000 N^2 - 2500 N^2)
F2 = sqrt(7500 N^2)
F2 = 86.60 N (rounded to two decimal places)
Therefore, the magnitude of one of the forces is approximately 86.60 N.