Question

Two force A and B at a point at right angles. If their resultant is 50N and their sum is 70N,

Answer 1

Taking a wild guess here, but it sounds like you're asked to find |A| and |B| given the magnitude of the resultant A + B, i.e. |A + B| = 50 N, and the sum of the individual magnitudes, |A| + |B| = 70 N.

Recall that the dot product of a vector with itself is equal to the square of that vector's magnitude,

A • A = |A|²

Then

|A + B|² = (A + B) • (A + B)

|A + B|² = (A • A) + 2 (A • B) + (B • B)

|A + B|² = |A|² + 2 (A • B) + |B|²

Since A and B are perpendicular to one another, their dot product is

A • B = 0

So it follows that

|A + B|² = |A|² + |B|²

2500 N² = |A|² + |B|²

Substitute |B| = 70 N - |A| and solve for |A| :

2500 N² = |A|² + (70 N - |A|)²

2500 N² = |A|² + 4900 N² - (140 N) |A| + |A|²

2 |A|² - (140 N) |A| + 2400 N² = 0

|A|² - (70 N) |A| + 1200 N² = 0

|A|² - (70 N) |A| = - 1200 N²

|A|² - (70 N) |A| + 1225 N² = - 1200 N² + 1225 N²

(|A| - 35 N)² = 25 N²

|A| - 35 N = ± 5 N

|A| = 35 N ± 5 N

so that |A| = 30 N or |A| = 40 N. If we fix |A| to be one of these, then |B| will have the other value.

So, the magnitudes are |A| = 30 N and |B| = 40 N.