- An object in equilibrium has three forces exerted on it. A 33-N force act at 90° from the x-axis and a 44-N force act at 60°. What are the magnitude and direction of the third force

The magnitude and direction of the third force is;

F3 = 74.44 N

θ3 = 72.81° in the south direction

We are given;

F1 = 33 N

F2 = 44 N

θ1 = 90°

θ2 = 60°

Let the third force be F3 which will serve as the resultant

Let's first find the x and y component of the forces.

F1x = F1 cos θ1

F1x = 33 × cos 90

F1x = 0 N

F1y = F1 sin θ1

F1y = 33 × sin 90

F1y = 33 N

F2x = F2 cos θ2

F2x = 44 × cos 60

F2x = 22 N

F2y = F2 sin θ2

F2y = 44 × sin 60

F2y = 38.11 N

Thus, the resultant of the x component is;

F3x = F1x + F2x

F3x = 0 + 22

F3x = 22 N

The resultant of the y component is;

F3y = F1y + F2y

F3y = 33 + 38.11

F3y = 71.11 N

Thus, magnitude of resultant of the F3 force is;

F3 = √((F3x)² + (F3y)²)

F3 = √(22² + 71.11²)

F3 = 74.44 N

The direction of the resultant of F3 is;

θ3 = tan^(-1) F3y/F3x

θ3 = tan^(-1) 71.11/22

θ3 = 72.81° in the south direction

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The magnitude of third force is 74.4 N and direction of third force  is 72.8 degrees South.

Explanation:

Let F1, F2 and F3 are three forces exerted on an object.

We have to find the direction and magnitude of third force i.e F3.

Now,

x-component of  resultant

y-component of resultant

South

Hence, the magnitude of third force is 74.4 N and direction of third force  is 72.8 degrees South.

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