I need HELP Pleassseeee!!!!
Firstly - they state the ring is 'smooth. This means the tension is the same in AR as in BR. THe diagram shows all the forces acting on the apex r: |dw:1393014306365:dw| You can derive some other angles from the geometry: 3 angles = theta |dw:1393014572640:dw| So oyu can resolve the 2 tensions vertically and they must balance the weight resolve them horizontally - they balance the pulling force Sart there - see how you do...
I pretty much got that part, but how do you resolve the angles
In the answer they had Tcosθ + Tsinθ = 15.5 –Tcosθ + Tsinθ = 8.5
How did they arrive at that?
do you know how to resolve a force into components?
Yes, but this one has been quiet tricky
so what is the horizontal component of the tension in AR?
Tcosθ
and what is the horizontal component of the tension in BR?
BR is where I found it tricky :(
you are used to having the angle to resolve in but in this case you have a right angle triangle between BR, the vertical, and the horizontal so the force is at 90-theta to the horizontal can oyu work from that?
okay and cos(90-θ) is sinθ
I see it now :) thanks
The 2 equations come from resolving horiz and vertical They are independent simultaneous equations so the answers drop out from them
Could you rephrase that?
You have correctly got the cos(90-theta) = sin(theta) So the first equation you gave above is from resolving horizontal components The second equation will come from a similar process in resolving vertical forces The 2 equations are independent (since horiz and vert forces do not affect each other) They are also simultaneous (since both are true from the same system) So oyu can solve the two equations by whatever method you know for simultaneous equations. That gives you the values for TsinTheta, and TcosTheta as given in the question Find TanTheta from those, and hence theta
As a general point - if you resolve forces or vectors, then if 1 axis is Fcos Theta then the other is Fsin Theta
Thanks Thanks I got that :)
Well done. :-)
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