T2

T 1 = 400 N

    57° 30°

75 kg.

y T2

T1 400N

57° 30°

x

Fg -735N

T1x = T1 COS θ

T1x = (400N) (COS 30°)

T1x = 346.4N

T1y = T1 SIN θ

T1y = (400 N) SIN 30°

T1y = 200 N

y

T2y

x

T1y 200 N T2x T1x 346.4 N

Fg – 735 N

If you used the “x” direction

T1x = T2

346.4 N = (T2)(cos 57°)

346.4 N/(cos 57°) = T2

636 N = T2

If you used the “y ” direction

Fy net = Fs + Fg

substitute Ty net = T1y + T2y + Fg

0 = 200 N + T2y + -735 N T2y = 535 N

THEN

If you used the “y ” direction

εFy = may

T1y + T2y + Fg = (75 kg)(0 m/s2) T1y + T2y + Fg = 0

(T2) (sin 57°) = T2y

(T2) (sin 57*) = 535 N T2 = 535N/(sin 57°) T2 = 637 N