3 forces act on an object. A force of 100N acts towards the North, a force of 240N acts toward east, and a force of 150N acts at S20E.

Question

determine the magnitude and direction of the resultant force of these three forces. 

Include a diagram in your solution. 

I need help with this ASAP please! 

Answer 1

The tail of each vector share a common point.  That point represents the object.

 

 

First, we need to add up all of the x components and add up all of the y components of the 3 vectors.

 

For 100 N due north

 

x-component is 0 N.

y-component is 100 N.

 

 

For the 240 N due east

 

x-component is 240 N.

y-component is 0 N.

 

 

These two vectors are not difficult to breakdown to their components since they are horizontal and vertical.

 

For the 150 N due 20 degrees southeast:  Southeast is in the fourth quadrant.  So we subtract 20 from 360.  That will be the angle we use.

 

x-component = 150cos(340) = 140.95 N

y-component = 150sin(340) = -51.30 N

 

 

Now we add the x components and y components.  This will give us the resultant R_x and R_y.

 

R_x = 0 N + 240 N + 140.95 N

    = 380.95 N

 

R_y = 100 N + 0 N - 51.30 N

     = 48.70 N

 

 

Use the component resultants to find the resultant R of the entire vector.  We use the Pythagorean theorem.

 

R = √(R_x² + R_y²)

   = √(380.95² + 48.70²)

   = 384.05 N

 

This is the resultant of the combine vectors.

 

 

To find the direction, take the inverse tangent of R_y over R_x.

 

θ = tan^-1(R_y / R_x)

   = tan^-1(48.70 / 380.95)

   = 7.29

 

The direction is 7.29 degrees due northeast.