Question

Three vector force F1, F2 and F3 act on a particle of mass m= 3.80kg as show fig. (I) calculate the magnitude and direction of the net force acting on the particle.

Answer 1

see figure, here Three vector force F1, F2, and F3 act on a particle of mass , m = 3.8kg.

components of F1 :

component of F1 along x -axis = -80cos35° [ negative, because it is acting along negative direction of x-axis]

component of F2 along y-axis = 80sin35°

similarly, components of F3 :

component of F3 along x-axis = 40cos45°

component of F3 along y -axis = -40sin45°

now, net force acting on mass = F1 + F2 + F3

= (-80cos35 i + 80sin35°j) + 60i + (40cos45° i - 40sin45° j)

= (-80cos35° + 60 + 40cos45°)i + (80sin35° - 40sin45°) j

[ cos35° = 0.82 , sin35° = 0.57 , sin45° = cos45° = 1/√2 = 0.707]

now, net force = (-80 × 0.82 + 60 + 40 × 0.707)i + (80 × 0.57 - 40 × 0.707)j

= 22.68i + 17.32j

magnitude of force = 28.536 N

magnitude of force = 28.536 N angle made by net force with x-axis is , α = tan^-1(17.32/22.68) ≈ 37.37°