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Physics 14 Online
OpenStudy (anonymous):

A plank of length L=1.800 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at the other end as shown. The cable is attached at a height H=1.70 m above the pivot and the plank's CM is located a distance d=0.600 m from the pivot. Calculate the tension in the cable.

OpenStudy (anonymous):

So lets start by finding the length of the cable. d= \[\sqrt{1.8^2+1.7^2}\] d=2.48 Now let T represent Tension mgd1-T(H/d2)l 4(9.81)(.6)-T(1.7/2.48)1.8=0 23.544=1.2T T=19.62N I'm pretty sure this is how you go about solving for tension!

OpenStudy (anonymous):

thats what I got too! But its telling me its wrong :(

OpenStudy (anonymous):

Oh wait I think we just calculated wrong! Thank you so much for explaining to me! :)

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