3.
T2
T 1 = 400 N
57° 30°
75 kg.

4. y
T2
T1 400N
57° 30° x
Fg -735N

5. T1x = T1 COS θ
T1x = (400N) (COS 30°)
T1x = 346.4N

6. T1y = T1 SIN θ
T1y = (400 N) SIN 30°
T1y = 200 N

7. y
T2y
x
T1y 200 N
T2x T1x 346.4 N

8. If you used the “x” direction
T1x = T2
346.4 N = (T2)(cos 57°)
346.4 N/(cos 57°) = T2
636 N = T2

9. 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

10. 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