Question

S2-602 Spiderman casts two web lines HE and HD across two buildings as shown in the figure below. In doing so, he applied a force of 150 N on line HE and 250 N on line HD pulling from point H Determine: a) The resultant, R, of the two forces acting at the point H b) The angle, , between the two web lines at H c) The moment of the resultant, R, about the point B, and d) The moment of the resultant, R, about the line BM y 15 m H K 50 m 40 m N E L M 20 m C X 100 m 50 m A COORDINATES: B(30,0,50) E(30,100,50) D(30,115,0) H(70,40,20) M05,0-20) 30 m 40 m 25 m

Answer 1

Tension T uE Magnitude of tension, 150 N IHF The force is directed from whose coordinates are as follows: H to E -20 m 70 40 H E ( 100 50 30 Therefore, the position vectors are: 70 i+ 40 j 30 i 100i + он 3 ОЕ %3 20 k 50 k ОE - он Hence, HE = -40 i 60 70 k The magnitude of the vector is obtained as: 40 2 + 602 70 E 100.499 m The unit vector along the direction of this vector is: НЕ/ НЕ -0.398 i + 0.697 k 0.597 Hence, the tension vector can be written as: Т HЕ 3 Т НЕ И 59.700 i НЕ 104,550 k 89.550 N - Tension TH D Magnitude of tension, 250 N HO = The force directed from H to D whose coordinates are as follows: H( -20) 70 40 m. 0 D( Therefore, the position vectors are: 115 30 70 i40 - 30 i115 j + 20 k OH= O D= 0 lo но 3 оD - он Hence. = 40 i + 75 20 k The magnitude of the vector is obtained as: H D 40 75 20 = 87.321 m The unit vector along the direction of this vector is HD HD Пио 3 0.230 k -0.458i + 0.859 Hence, the force vector can be written as: Тирй Тир 3 но -114.500 i + 214.750j 57.500 k N

a) RESULTANT OF THE TENSIONS Therefore, the resultant of the two tensions is obtained by vector addition: R = ТHE+ THD -174.200 i + 162.050 k 304.300 N The magnitude of the vector is obtained as: 386.270 R = N b) The angle between the two vectors, ТHE THD cos a TNE TO a cos 32078.138 37500 = 31.194

c]Moment of R about B -174.200 i 162.050 k 304.300 j N = B MOMENT OF R ABOUT B are The coordinates of в ( ов - m 50 k 30 0 50 Hence, 30 i + is obtained as: Therefore, moment of R about B M = овх R k i 30 0.000 50.000 304.300 -174.200 162.050 -15215,000 i 13571.500 9129.000 k N m d) В AND M МОМENT OF R ABOUT AXIS THROUGH The coordinates of M are M ( )m 95 -20 OM O B B M Hence, 70 k 0 j 65 i = The magnitude of the vector is obtained as: B M 0 -70 95.525 m The unit vector along the direction of this vector is вм / Вм 0.680 i 0.000 0.733 B M Hence, moment of R about axis - Ма . й в м -17037.757 N.m