Question

how would you solve the last problem?

R145 TE0,8SN Ag Ts= 1.21 N Ap A, CasB-t1-Cos-CY Measure the y and a angles for each tension force and then calculate the angle B using the Pythagorean relationship: 1- cos2 (a) + cos2 (B) + cos2 (y). Make sure to keep track of the octant, as this sets the sign convention. (Please attach any work you do to get these values, check your work before you leave). Y T3-1.3 N TI-D.6 Magnitude (Newtons) 61 64.5 135 1029 30 45 Y Octant V (9 pts) (s)5u-1(9t),S0-

Answer 1

Tx = T coso sina Ty = Tsino sind Tz = T cosa NOW Ty Tx = Tsino sing The Toso sing = tano w and, T +Ty = + 2 sinop (costo + sin o) = T² sina Tsing = S try? Tsing = T2? + Tyl in Tz Tz Tesa Ja Ty tom 4 for 7 tension, Tixo - 0.473 Tiy=- 2012 N Tiz =0.3 44 N. Ta tam = Tiy -00123 = 0.26 Tom 0-473- ** 0,= tanto.26.) = 14,57° - tan & Ten + Tigt (-0.473)2 + (-0.123) 2 0.344 4, = 54.860 T = To coso, sing it, sing, sing 3 + Ti coop in = 0.6 [cos (14.57°) sin (54-86°) + sin (14.52°) Son ( 54.86") it cos (57.86) ]

Similarly, for T2 , x = 0.0314N, Toy = -0.636N, Tzz = 0.636 N. Bu town a tool (2004). Pzstant Teata mos) ston5 (0.021)** 60.636) 2 09 - - 87.17 Tza o. 636 mm hocilorsnog 2 02 ²450 T2 = ta loss of sind, i taa sind, sind it Tz losoa u = 0.9 [cos (-87012") Sin 45 it. sin (-87:97°) sin 450 & losaso û ] . 10 10 orari