Vectors.........Physics Q?

Question

1.) The following forces act on an object resting on a level, frictionless surface. 10 N to the north, 20 N to the east, 10 N at an angle 40 degrees south of east, and 20 N at an angle 50 degrees west of south. Find the magnitude and direction of the resultant force acting on the object.

Answer 1

To do vector addition, break down the vectors into their components.

A and B are already broken down.

So for C, the x component is 10N cos (-40 degrees)
The y component is 8N sin (-49 degrees)

The -40 degrees uses the convention that all angles are measured counterclockwise from +x (EAST)

The last angle is -140.

Then add the components.

Use pythagorus to get the total magnitude.

Use arctan (y/x) to get the angle. Use your common sense to resolve the ambiguity in the arctan function.

Good luck!

Answer 2

Ok, here's how you do it, first you draw your free body diagram drawing each vector as an arrow with it's direction.

Then you asign a coordinate axis to the drawing, in my case, i supposed that the north is going to be the y axis and the east is going to be the x axis.

now you add all the forces in the x axis, being carefull with the orientation.

Sum of forces on x = 20N + 10N cos40 - 20N sin 50

Notice that the 20N is going to be negative because it's directed towards the west.

Sum of forces on x = 12.34N

Now you add the forces on the y axis.

sum of forces on y = 10N - 10Nsin40 - 20Ncos50

sum of forces on y = -9.28N

the negative means that it's directed towards south.

Now you get the resultant force

F = sqrt(12.34N**2 + (-9.28N)**2) = 15.44N

by ** I mean that the number is squared and sqrt is the squared root.

now you need to get the direction, you do it this way

angle = tan-1(9.28/12.34) = 36.9º south of east.

Hope this helps.