Three forces of magnitudes f1=4.0n, f2=6.0n, and f3=8.0n are applied to a block of mass m=2.0kg, initially at rest, at angles shown on the diagram. (figure 1) in this problem, you will determine the resultant (net) force by combining the three individual force vectors. all angles should be measured counterclockwise from the positive x axis (i.
e., all angles are positive).

The resultant force can be found by finding the sum of the vectors of the individual forces

The resultant (net) force vector and magnitude of the resultant of the three forces in the question are

The reason the above values re correct is as follows:

The given forces, are;

F₁ = 4.0 N, direction; 25° North of East

F₂ = 6.0 N, direction; 325° counterclockwise measured from the x-axis = 35° South of East

F₃ = 8.0 N, direction; 180° counterclockwise = West

The representation of the forces in vector format is as follows;

= 4 × cos(25°)·i + 4 × sin(25°)·j

= 6 × cos(325°)·i - 6 × sin(325°)·j

= -8 × cos(180°)·i + 8 × sin(180°)·j

The resultant force, is given as follows;

= + +

∴ = (4 × cos(25°) + 6 × cos(325°) - 8 × cos(180°))·i + ((6 × sin(25°) - 6 × sin(325°) +  8 × sin(180°))·j

Which gives;

= (4 × cos(25°) + 6 × cos(325°) - 8)·i + ((6 × sin(25°) + 6 × sin(325°))·j

= 0.54·i - 1.75·j

The resultant force,  = 0.54·i - 1.75·j

The direction of the resultant force, θ = arctan((- 1.75)/0.54) ≈ - 72.85° ≈ 287.15° Counterclockwise from the positive x-axis direction

The magnitude of the resultant force, = √(0.54² + 1.75²) ≈ 1.83142

The magnitude of the resultant force, ≈ 1.83142 N

Learn more about force vectors here:

link

the answer is 12 as i may remember

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