Sum of magnitude of two forces is 20 N and magnitude of their resultant is 10 N. If resultant is perpendicular to the smaller force find values of both forces.

Options

A. 5.5 N, 14.5 N
B. 7.5 N, 12.5 N
C. 15 N, 5 N
D. 12 N, 8 N

Answer

To solve the problem, we will denote the two forces as \( F_A \) and \( F_B \). We know the following:

1. The sum of the magnitudes of the two forces is 20 N:
\[ F_A + F_B = 20 \quad \text{(1)} \]

2. The magnitude of the resultant force \( F_R \) is 10 N:
\[ F_R = 10 \quad \text{(2)} \]

3. The resultant \( F_R \) is perpendicular to the smaller force, which we will assume to be \( F_B \).

Using the Pythagorean theorem, since \( F_R \) is perpendicular to \( F_B \), we can write:
\[ F_R^2 = F_A^2 - F_B^2 \quad \text{(3)} \]

Substituting the known value of \( F_R \) from equation (2) into equation (3):
\[ 10^2 = F_A^2 - F_B^2 \]
\[ 100 = F_A^2 - F_B^2 \quad \text{(4)} \]

Now, we can express \( F_A^2 - F_B^2 \) in terms of \( F_A + F_B \) and \( F_A - F_B \). We can rewrite equation (4) as:
\[ F_A^2 - F_B^2 = (F_A + F_B)(F_A - F_B) \]

From equation (1), we know \( F_A + F_B = 20 \). Let’s denote \( F_A - F_B \) as \( x \):
\[ F_A^2 - F_B^2 = 20 \cdot x \quad \text{(5)} \]

Now we can set equations (4) and (5) equal to each other:
\[ 20x = 100 \]
\[ x = \frac{100}{20} = 5 \quad \text{(6)} \]

Now we have \( F_A - F_B = 5 \) (from equation (6)). We can now solve the system of equations:
1. \( F_A + F_B = 20 \) (equation (1))
2. \( F_A - F_B = 5 \) (equation (6))

Adding these two equations:
\[ (F_A + F_B) + (F_A - F_B) = 20 + 5 \]
\[ 2F_A = 25 \]
\[ F_A = \frac{25}{2} = 12.5 \, \text{N} \quad \text{(7)} \]

Now substituting \( F_A \) back into equation (1) to find \( F_B \):
\[ 12.5 + F_B = 20 \]
\[ F_B = 20 - 12.5 = 7.5 \, \text{N} \quad \text{(8)} \]

Thus, the magnitudes of the two forces are:
\[ F_A = 12.5 \, \text{N} \quad \text{and} \quad F_B = 7.5 \, \text{N} \]

### Summary of the Solution:
- Magnitude of Force A (\( F_A \)) = 12.5 N
- Magnitude of Force B (\( F_B \)) = 7.5 N