Question

Force Problem: Four concurrent forces are applied to a 5 Kg mass: 50 N North, 25 N East, 35 N South, and 35 N West.

Answer 1

A) Net horizontal force:
Fx= -35 + 25 + -10N

Net vertical force
Fy= 50 - 35 = 15N

Net force
F^2 = fX^2 +Fy^2
F= 18N

B.) Acceleration of system
a=18/5=3.6m/s^2

C). Equilibriant force must be equal to the net force above
F1= 18N

D.) Acceleration becomes half
a1= a/2 =1.8m/s^2

E.) Equilibrant force also doubles
F" = 2F' = 36N

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