three vector forces f1,f2 and f3 acts on particle of mass m=3.80kg
1-calculate the magnitude and the direction of the net force acting on the particle.
2-calculate the particle acceleration.
3-if an additional stabilizing force f4 is applied to create an equilibrium condition is a resultant net force of zero,what would be the magnitude and direction of f4.
Answers
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°
2. acceleration of the particle , a = F/m
= 28.536/3.8 ≈ 7.5 m/s²
3. if an additional stabilizing force f4 is applied to create an equilibrium condition is a resistant net force is zero.
then according to Newton's 3rd law,
magnitude of f4 = 28.536 N
and direction of f4 just opposite to direction of 28.536N force .
Answer:
Picture is the answer
Explanation:
