Question

find the magnitude and direction of (A-B), if A is 50.00 N at 30.0 degrees and B is 50.00 N at 0.00 degrees?

Answer 1

Given that

A = 50.0 N at 30 deg above x-axis

Components of vector A will be:

A = Ax i + Ay j

Ax = 50.0*cos 30.0 deg = 43.3 N

Ay = 50.0*sin 30.0deg = 25.0 N

B = 50.0 N at 0.00 deg with x-axis

Components of vector B will be:

B = Bx i + By j

Bx = 50.0*cos 0 deg = 50.0 N

By = 50.0*sin 0 deg = 0 N

So,

R = A - B

R = (Ax - Bx) i + (Ay - By) j

R = (43.3 - 50.0) i + (25.0 - 0.0) j

R = -6.7 i + 25.0 j

Now magnitude of Vector R will be:

|R| = sqrt ((-6.70)^2 + (25.0)^2)

|R| = 25.88 N = magnitude

Direction = arctan (Ry/Rx)

Direction = arctan (25.0/(-6.7)) = -75.0 degrees = 75.0 deg above the -ve x-axis

Direction = 180 - 75 = 105 deg above the +ve x-axis

Let me know if you've any query.