A particle is acted upon two forces given by the vector equations,

F1=1.5i-0.7j and
F2=-3.1i+3.6j,

with values measured in Newtons(N). Calculate the resultant force, in N, acting on the particle, giving your answer to 3 dec places.

The resultant vector is

-1.6i +2.9j
and its magnitude is
sqrt[1.6^2 + 2.9^2]= 3.31 N

-2.3i+4.5j

To calculate the resultant force acting on the particle, you need to find the vector sum of the two forces (F1 and F2).

Step 1: Add the x-components of the forces together:
Fx = F1x + F2x
= 1.5 + (-3.1)
= -1.6

Step 2: Add the y-components of the forces together:
Fy = F1y + F2y
= -0.7 + 3.6
= 2.9

Step 3: Combine the x and y components to get the resultant force vector:
Fresultant = Fx i + Fy j
= -1.6i + 2.9j

Step 4: Calculate the magnitude of the resultant force using the Pythagorean theorem:
Magnitude of Fresultant = sqrt((Fx)^2 + (Fy)^2)
= sqrt((-1.6)^2 + (2.9)^2)
≈ sqrt(2.56 + 8.41)
≈ sqrt(10.97)
≈ 3.314

Therefore, the resultant force acting on the particle is approximately 3.314 N, when rounded to 3 decimal places.