1. Calculate the tension in the string T in each of the following situations. In all calculations ignore the mass of the beam. 2. Moment=F X Perpendicular Distance and Basic Trigonometry 3.I have tried 200N X 0.5m=100 100/Cos30 Help is much apperchated.
I understand the left side of this equation: the clockwise torque due to the 200 N force. The right side should be the counter-clockwise torque due to the tension of the cable. Correct this equation, rewriting the right hand side in terms of T (tension). Then solve for T.
I think i have found the solution by calculating the "opp" tan30 X 0.5 = 0.2886 from there onward we do 200 X 0.5= 100/0.2886 =346.5/cos 30 = 400 N However now I am stuck on an similar equation of
Don't really understand what you did there with tangents and cosines. All you need is: 0.5*200 = 0.5*T*sin(30) which gives you T = 400. Solve it in the same way. The only difference is the moment arm associated with the 200 N force. I strongly suggest you adopt a systematic approach, like the one I used.
But why would you use sine30 wouldnt that work out the "opp"?. Due to the fact the T is on the hypoteanus.
You want the component of T perpendicular to the beam, which in this case means [itex]T\sin {30}[/itex].
But why would T need to become perpendicular ? Also for question 2 how am I ment to work it out with out the given length ?
Because you are finding the torque (moment) it generates about the pivot point. Do you understand the definition of moment?
That's correct. But realize that you don't need the length of the beam: You can just use L or L/2. (The actual length doesn't matter.)
But doesnt the formular state Motion=Force * perpendicular distance ? This would mean the formula is 200*L = L*T*sin30 correct ?
This is for the second problem though, wouldnt the Motion be different due to the force being applyed to the mid-point? I seem to of gotten T=400 in my previous post.
These calculations apply to the first problem. For the second problem, the method is exactly the same: Set clockwise and counter-clockwise torques equal. Of course the torque exerted by the 200 N force is different in the second problem, since it's applied at the midpoint instead of at the end.