The magnitude and direction of two forces acting on an object are 50 pounds, S68°E, and 110 pounds, N54°E, respectively.

Question

Find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.

Answer 1

Assuming the notation means start at South then move 68 degrees East.

The best way to do this is to look at the NS and EW components separately.

The 50 pound force is going at an angle down and to the right.

Down is negative and right is positive.

By placing the vector as the hypotenuses of a right triangle we can see

the distance can be broken up into two components (vertical and horizontal)

These can be found with trig.

50 cos(68) = vertical = 18.73 down

50 sin(68) = horizontal = 46.36 right

Doing the same for the other force we get.

110 cos(54) = vertical = 64.66 up

110 sin(54) = horizontal = 88.99 right

Adding up all vertical and horizontal we get (remember down and left are negative)

-18.73 + 64.66 = 45.93 up

46.36 + 88.99 = 135.35 right

The magnitude is the hypotenuse.

sqrt(45.93² + 135.35² ) = 142.93 pounds.

Direction is acos(45.93/142.93) = 71.26 Remember this is North then 71.26 degrees East.