Comment Step 3 of 7 A As the no horizontal force acting on the load, the net horizonta ΣΕ = 0 T, cos 50.0°-T cos 40.0º =

[answerhappygod]

1 (62.92 KiB) Viewed 48 times

Comment Step 3 of 7 A As the no horizontal force acting on the load, the net horizonta ΣΕ = 0 T, cos 50.0°-T cos 40.0º = 0 T, cos 50.0º = 7 cos 40.0° 7) = 1.197 At equilibrium state of the load, the net vertical force on the load F, = 0 7, sin 40.0° +7, sin 50.0° - 7) = 0 T, sin 40.0° +T, sin 50.0º = 7 Substitute 1.197, for T, Ti sin 40.0° +(1.197,)sin 50.0º = T,