Question

CHW 3 Chs 2.8 and 2.9 7of7 Problem 2.129 Determine the magniude of the projęcted component of the 6 kN fonce acting along the xis IC of the pipe Suppose that a- 29 m and figures and include the appropriate units 1 of 1 nxel 600 2 m 8

Answer 1

answer ) first of all we have to find the position vectors

A=0i-2j+0k

B=0i+0j+0k

C=2.9i+3.8j-1k

D=7.9i+0j+0k

now

BA=A-B=0i-2j+0k

magnitude of BA=$\sqrt{}$0²+(-2)²+0²=2

BC=C-B=2.9i+3.8j-1k

magnitude of BC=$\sqrt{}$2.9²+3.8²+(-1)²=4.88

so unit vector along BC=BC/magnitude of BC=2.9i+3.8j-1k/4.88=0.594i+0.779j-0.205k

CD=D-C=5i-3.8j+1k

magnitude of CD=$\sqrt{}$5²+(-3.8)²+(1)²=6.36

so unit vector along CD=CD/magnitude of CD=5i-3.8j+1k/6.36=0.786i-0.597j+0.157k

now for cartesian form of the vector we have

F=6kN*unit vector CD=6kN*(0.786i-0.597j+0.157k)=4.716i-3.582j+0.942k

now the component along BC=F* unit vector BC=4.716i-3.582j+0.942k*0.594i+0.779j-0.205k=2.801-2.790-0.193=-0.182kN=-182N

we only want magnitude so the answer is 182 N