Two forces whose resultant is 10N are perpendicular to each other. If one of them makes an angle of 60 degree with the resultant,calculate its magnitude

Yes

To solve this problem, we can use vector addition and trigonometry. Let's assume that the magnitude of the first force is F1 and the angle it makes with the resultant is 60 degrees.

We know that the two forces are perpendicular to each other, so the angle between them is 90 degrees.
Using the Pythagorean theorem, we can express the magnitude of the resultant force as follows:

Resultant force (R)^2 = F1^2 + F2^2

Since the forces are perpendicular, we can use trigonometry to express the magnitude of the resultant force:

R = F1 * sin(60 degrees) = F1 * sqrt(3) / 2

Now, we can substitute this expression for R into the equation for the magnitude of the resultant force:

(F1 * sqrt(3) / 2)^2 = F1^2 + F2^2

3F1^2 / 4 = F1^2 + F2^2

Rearranging this equation, we get:

F1^2 = 4/3 (F1^2 + F2^2)

3F1^2 = 4F1^2 + 4F2^2

Substituting the given information, we have:

3(F1^2) = 4(F1^2) + 4(10^2)

3F1^2 = 4F1^2 + 400

F1^2 = 400

F1 = sqrt(400)

F1 = 20 N

Therefore, the magnitude of the force that makes an angle of 60 degrees with the resultant is 20 N.

To solve this problem, we can use vector addition and trigonometry.

Let's assume that the two forces are \(F_1\) and \(F_2\), with \(F_1\) making an angle of 60 degrees with the resultant. We are given that the resultant force is 10 N.

Since the two forces are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant force:

\((F_1)^2 + (F_2)^2 = (10 N)^2\)

Now, let's focus on \(F_1\) and find its magnitude. Since \(F_1\) makes an angle of 60 degrees with the resultant, we can use trigonometry. We can use either sine or cosine, but let's use cosine since it relates to the adjacent side.

\(\cos(60) = \frac{F_1}{10 N}\)

Simplifying, we have:

\(F_1 = 10 N \cdot \cos(60)\)

Now, we can calculate the value for \(F_1\):

\(F_1 = 10 N \cdot \cos(60) = 10 N \cdot \frac{1}{2} = 5 N\)

Therefore, the magnitude of \(F_1\) is 5 N.

If you draw the diagram, you can see that

|v-u|/|u| = tan60° = √3