Question

A (O.O.ZA) D (-5, 10, 0) B (15.-10,0) C (20, 25.0) COLLAPSE IMAGES F = 104N Fo=220N F;= 213N Zx = 41 m Best Score: 1.00/4 Current Score: 0.0074 Find the magnitude of the resultant force acting on the pole at height ZA N ENTER 3 tries remaining 1 points possible Calculate the directional cosine cos(0) ENTER 3ries remaining poids) sossible Calculate the directional cosine cos(5) ENTER 3 tres remaining 1 points) Bossible Calculate the directional cosine cos(V) ENTER 3 tres remaining points possible

Answer 1

Tension TAB Magnitude of force, TAB = 104 N The force is directed from A to B whose coordinates are as follows: Al 0, 0, 41 ) m B 15 -10, 0) m Therefore, the position vectors are: O A = 0 + 0 + 41 k OB = 151 - 10 ] + Ok Hence, AB - 0B - OA = 15 1 - 10 - 41 k The magnitude of the vector is obtained as: A B = /15? + -10? + -412 = 44.788 m The unit vector along the direction of this vector is: WA = A B/ AB = 0.335 i - 0.223 - 0.915 k Hence, the force vector can be written as: TAB - TAB AB = 34.840 i - 23.192 j - 95.160 k N Tension TAC Magnitude of force, TAC = 220 N The force is directed from A to C whose coordinates are as follows: A 0, 0, 41 ) m c( 20 25, 0) m Therefore, the position vectors are: O A = 0i + 0 + 41 k OC = 20 i + 25j + Ok Hence, Ася ос- оА = 20 i + 25 j - 41 k The magnitude of the vector is obtained as: AC = / 20? + 25 + -41? = 52.019 m The unit vector along the direction of this vector is: ÑAC = AC/ AC = 0.384 | + 0.481 ) - 0.788 k Hence, the force vector can be written as: TAC = TAN AC = 84.480 i + 105.820 - 173.360 k N

Tension TAD Magnitude of force, TAD = 213 N The force is directed from A to D whose coordinates are as follows: AT 0, 0, 41 ) m D -5, 10, 0) m Therefore, the position vectors are: O A = 0 + 0 + 41 k OD = -51 + 10 + Ok Hence, AD= OD OA = -51 + 10 41 k The magnitude of the vector is obtained as: AD=/ -5° +10° + -41 - 42.497 m The unit vector along the direction of this vector is: WAD = AD/AD = -0.118 i + 0.235 - 0.965 k Hence, the force vector can be written as: TAD = TAD AD = -25.134 i + 50.055 j - 205.545 k N RESULTANT OF FORCES Therefore, the resultant of the three forces is obtained by vector addition: R = TA8+ TAC + TAD = 94.186 i + 132.683 - 474.065 k N The magnitude of the vector is obtained as: R = 501.212 N 94.186/501.212) Its inclinations with respect to the axes are: cos a = 1 = 0.188 cos B = 1 = 0.265 cos y = = -0.946 132.683/501.212 ) -474.065/501.212)