Recent content by rickbusiness

nevermind, figured it out. for anyone doing a similar problem here's my sol'n Fxhinge=Tsin(theta) Fy=0=Tcos(theta)-mg+Fyhinge Fyhinge=mg-Tcos(theta) Fxhinge squared + Fyhinge squared = F hinge squared

Homework Statement A 35.8 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of theta=17.8° with respect to horizontal, what is the horizontal component of the force exerted by...

hahaha thanks!

Ax=Acosθ=47.2cos(54.3) = 27.5 Ay=Asinθ=47.2sin(54.3) = 38.3 Bx=Bcosθ=23.5cos(180-34.4) = -19.4 By=Bsinθ=23.5sin(180-34.4)= 13.3 Cx=Ccosθ=30.5cos(270)=0 Cy=Csinθ=30.5sin(270)=-30.5 Rx= 27.5-19.4+0=8.1 Ry=38.3+13.3-30.5=21.1R=√(8.12)+(21.12)=22.6 θ=tan-1(21.1/8.1)=68.9deg Ive tried...

http://postimage.org/image/gcb728073/ A=47.2 at 54.3deg above the x axis, B= 23.5 at 34.4 above the -x axis, C=30.5 along the -y axis. A+B+C=(Ax+Bx+Cx)+(Ay+By+Cy) = R R=√(Rx2+Ry2) = 22.63 θ=tan-1Ry/Rx = 68.9deg I just thought of the possibility that c may be at 90 deg...