Question

two forces F1 and F2 act on a 6.80-kg object. F1=30.0 N and F2 = 11.0 N. 

(a) Find the acceleration of the object for the configuration of forces shown in figure (a)

(b) Find the acceleration of the object for the configuration of forces shown in Figure (b). 

Answer 1

Part A )

m = 6.8 kg

F₁ = 30 N

F₂ = 11 N

F = ( F₁² + F₂² )^1/2

m a = ( 30² + 11² )^1/2

a = 31.95 / 6.8

a = 4.698 m/sec²

$\theta$ = tan^-1 ( 11/30 )

$\theta$ = 20.13^o

Part B )

F_1x = F₁ + F₂ cos60

= 30 + 11 X 0.5

= 35.5 N

F_1y = F₂ sin60 = 11 X 0.866 = 9.526 N

a = ( 35.5² + 9.526² )^1/2 / 6.8

a = 49.73 m/sec²

$\theta$ = tan^-1 ( 9.526/35.5 )

$\theta$ = 15.0207^o