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# M2 Statics - More Tension Problems Watch

1. On part c.. I'm confused with the markscheme for this question. They've replaced the light rod CD with the light rod CM, as described. However, they still have the length AC as 1m, and the angle ACM is 45 degrees, just as ACD was, is this correct, how did they work that out?

I tried using theta as the angle (as it's unknown) and xm for AC (as the length is also unknown). I then resolved vertically to get V = -Tcos(theta). V is the vertical component of the force at A, T is the thrust in the light rod, CM. I need to somehow show that V = 0 for the assumption to be true. I then resolved horizontally to get H = -Tsin(theta), where H is the horizontal component of the force at A. Note here that I had V facing north, and H facing east. Then took moments about A to get 40g = Tcos(theta) (after dividing by 1.5). This then led to V = -40g/cos(theta) x cos(theta) and therefore V = -40g. What have I done wrong?
2. (Original post by ViralRiver)

On part c.. I'm confused with the markscheme for this question. They've replaced the light rod CD with the light rod CM, as described. However, they still have the length AC as 1m, and the angle ACM is 45 degrees, just as ACD was, is this correct, how did they work that out?

I tried using theta as the angle (as it's unknown) and xm for AC (as the length is also unknown). I then resolved vertically to get V = -Tcos(theta). V is the vertical component of the force at A, T is the thrust in the light rod, CM. I need to somehow show that V = 0 for the assumption to be true. I then resolved horizontally to get H = -Tsin(theta), where H is the horizontal component of the force at A. Note here that I had V facing north, and H facing east. Then took moments about A to get 40g = Tcos(theta) (after dividing by 1.5). This then led to V = -40g/cos(theta) x cos(theta) and therefore V = -40g. What have I done wrong?
For the vertical components the equation at equilibrum
V+Tcos(theta)-40g=0
Horizontally
-H+T*sin(theta)=0
T*cos(theta)-40g*1.5=0
So V is the difference between 40g anf Tcos(theta) (is positive = facing north)

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Updated: December 21, 2010
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