Single force replacing two forces?

Question

Two forces, 80N and 100N acting at an angle of 60 degrees with each other, pull on an object. What single force would replace the two forces?

Answer 1

The interesting thing about this problem is how everyone got a different answer.

The Captain got it right, however; F = 156.2 N @ 26.3° from the 100 N force.  I checked it with a scale drawing after getting that answer myself.......

Answer 2

Well, if my math is right, and you should check it, it would be a single force of 91.652N acting at an angle of 19 degrees from the 100N force towards to 80N force.  So if the 80N and 100N forces were each 30 degrees from vertical, the 91.652N force would be 11 degrees from vertical towards the 100N force.

Answer 3

You do it with vectors and geometry

Establish that the 80N force is along one axis and we will center things on that

Resolve the 100N force into two vectors, one in the direction of the 80N force and one perpendicular to it.  This is done using Cosine.  Draw a right triangle with two unknown legs and the hypot being 100N and the angle at one end of 60 degees.  The "horizontal leg" will be the force along the 80N direction, the vertical leg will be that perpendicular to it.  The horizontal part of the 100N would be

Force = cosine (60 degrees) * hypotenuse

=0.5 * 100

=50

This is the portion of the force parallel to the 80N direction, add these together to get the full component in that direction of 130 N

The other portion would be

Force = sine (60 degrees) * hyp

=86.6

Now draw a new triangle to add the two vectors (the summed one and the new perpendicular one) and calculate the magnitude of the result vector using pythag

Magnitude = sq root (130^2 + 86.6^2) = 156.2 N

The angle it would be to the original 80N force would be

Tan angle = opposite/adj

angle = arctan (86.6/130)

angle = 33.7 degrees

You could also do it by geometery by drawing the 80N force horizontal and at its endpoint starting the 100N force at the 60 degree angle.  The sum would be a line connecting the start of the 80 with the end of the 100, I'm just to lazy to do the geometry

Answer 4

Forces are vectors so you just add them as vectors. Select an easy coordinate system to work in and in this case I think it is one with the x-axis along one of the vectors (I will pick the 100N one but it really doesn't matter which one). Now write each force as a vector using this coordinate system.

100N ...... F1 = 100i

80N ........ F2 = 80cos(60)i + 80sin(60)j

F2 = 40i + 40SQRT(3)j

F1 + F2 = F = 100i + 40i + 40SQRT(3)j

F = 140i + 40SQRT(3)j

The angle of the new force can be found from the components of the single force vector F. The angle will be between the two vectors and is measured from the 100N one.

Tan(Angle) = 40SQRT(3)/140 = 0.49487

Angle = 26.33 degrees

The magnitude of F will be the single force.

Magnitude = SQRT[(140)^2 + (40SQRT(3))^2]

Magnitude = 156.2 N

So the single force is 156.2N at an angle of26.33 degrees with respect to the 100N vector.

Answer 5

answer is the square root of 80^2+100^2-2x80x100xcos(60)