2012 ppt spiderman
- 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